Question
The function f is defined by f (x) = ex sinx , where x ∈ R.
(a) Find the Maclaurin series for f (x) up to and including the x3 term.
(b) Hence, find an approximate value for \(\int_{0}^{1}e^{x^{2}} sin(x^{2})dx.\)
The function g is defined by g (x) = ex cos x , where x ∈ R.
(c) (i) Show that g(x) satisfies the equation g ″(x) = 2(g′(x) – g(x)).
(ii) Hence, deduce that g(4) (x) = 2(g″′(x) – g ″(x)) .
(d) Using the result from part (c), find the Maclaurin series for g(x) up to and including the x4 term.
(e) Hence, or otherwise, determine the value of \(\lim_{x\rightarrow 0}\frac{e^{x}cos x – 1 – x}{x^{3}}.\)
▶️Answer/Explanation
Ans:
(a) METHOD 1
recognition of both known series
attempt to multiply the two series up to and including x3 term
Note: Condone absence of limits up to this stage.
Note: Accept working with each side separately to obtain -2ex sin x .
Note: Accept working with each side separately to obtain -4ex cos x .
Note: Do not award any marks for approaches that do not use the part (c) result.
Note: Condone the omission of +… in their working.
Question
Use l’Hôpital’s rule to determine the value of
lim \(\frac{2sinx-sin2x}{x^3}\) [6]
x → 0
▶️Answer/Explanation
Ans:
Question
Use l’Hôpital’s rule to find
\(lim_{x\rightarrow 1}\frac{cos(x^2-1)-1}{e^{x-1}-x}\)
▶️Answer/Explanation
Ans:
attempt to use l’Hôpital’s rule
= \(lim_{x\rightarrow 1}\frac{-2xsin(x^{2}-1)}{e^{x-1}-1}\)
Note: Award A1 for the numerator and A1 for the denominator.
substitution of 1 into their expression =\( _{0}^{0}\)hence use l’Hôpital’s rule again
Note: If the first use of l’Hôpital’s rule results in an expression which is not in indeterminate form, do not award any further marks.
attempt to use product rule in numerator
= \(lim_{x\rightarrow 1}\frac{-4x^{2}cos(x^{2}-1)-2 sin (x^{2}-1)}{e^{x-1}}=-4\)
Question
Use l’Hôpital’s rule to find \(lim _{x\rightarrow o}(\frac{arctan2x}{tan3x})\)
▶️Answer/Explanation
Ans
Question
Use l’Hôpital’s rule to find \(lim _{x\rightarrow o}(\frac{arctan2x}{tan3x})\)
▶️Answer/Explanation
Ans
Question
Calculate the following limit
\(\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} – 1}}{x}\) .[3]
Calculate the following limit
\(\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + {x^2})}^{\frac{3}{2}}} – 1}}{{\ln (1 + x) – x}}\) .
▶️Answer/Explanation
Markscheme
\(\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} – 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{2^x}\ln 2}}{1}\) M1A1
\( = \ln 2\) A1
[3 marks]
EITHER
\(\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + {x^2})}^{\frac{3}{2}}} – 1}}{{\ln (1 + x) – x}} = \mathop {\lim }\limits_{x \to 0} \frac{{3x{{(1 + {x^2})}^{\frac{1}{2}}}}}{{\frac{1}{{1 + x}} – 1}}\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{3x{{(1 + {x^2})}^{\frac{1}{2}}}(1 + x)}}{{ – x}}\) A1A1
\( = – 3\) A1
OR
\({(1 + {x^2})^{\frac{3}{2}}} – 1 = 1 + \frac{3}{2}{x^2} + \ldots – 1 = \frac{3}{2}{x^2} + \ldots \) M1A1
\(\ln (1 + x) – x = x – \frac{1}{2}{x^2} + \ldots – x = – \frac{1}{2}{x^2} + \ldots \) M1A1
Limit \( = – 3\) A1
[5 marks]
Question
By evaluating successive derivatives at \(x = 0\) , find the Maclaurin series for \(\ln \cos x\) up to and including the term in \({x^4}\) .
Consider \(\mathop {\lim }\limits_{x \to 0} \frac{{\ln \cos x}}{{{x^n}}}\) , where \(n \in \mathbb{R}\) .
Using your result from (a), determine the set of values of \(n\) for which
(i) the limit does not exist;
(ii) the limit is zero;
(iii) the limit is finite and non-zero, giving its value in this case.
▶️Answer/Explanation
Markscheme
attempt at repeated differentiation (at least 2) M1
let \(f(x) = \ln \cos x\) , \(f(0) = 0\) A1
\(f'(x) = – \tan x\) , \(f'(0) = 0\) A1
\(f”(x) = – {\sec ^2}x\) , \(f”(0) = – 1\) A1
\(f”'(x) = – 2{\sec ^2}x\tan x\) , \(f”'(0) = 0\) A1
\({f^{iv}}(x) = – 2se{c^4}x – 4se{c^2}xta{n^2}x\) , \({f^{iv}}(0) = – 2\) A1
the Maclaurin series is
\(\ln \cos x = – \frac{{{x^2}}}{2} – \frac{{{x^4}}}{{12}} + \ldots \) M1A1
Note: Allow follow-through on final A1.
[8 marks]
\(\frac{{\ln \cos x}}{{{x^n}}} = – \frac{{{x^{2 – n}}}}{2} – \frac{{{x^{4 – n}}}}{{12}} + \ldots \) (M1)
(i) the limit does not exist if \(n > 2\) A1
(ii) the limit is zero if \(n < 2\) A1
(iii) if \(n = 2\) , the limit is \( – \frac{1}{2}\) A1A1
[5 marks]
Question
Find the general solution of the differential equation \((1 – {x^2})\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 1 + xy\) , for \(\left| x \right| < 1\) .
(i) Show that the solution \(y = f(x)\) that satisfies the condition \(f(0) = \frac{\pi }{2}\) is \(f(x) = \frac{{\arcsin x + \frac{\pi }{2}}}{{\sqrt {1 – {x^2}} }}\) .
(ii) Find \(\mathop {\lim }\limits_{x \to – 1} f(x)\) .
▶️Answer/Explanation
Markscheme
rewrite in linear form M1
\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \left( {\frac{{ – x}}{{1 – {x^2}}}} \right)y = \frac{1}{{1 – {x^2}}}\)
attempt to find integrating factor M1
\(I = {e^{\int {\frac{{ – x}}{{1 – {x^2}}}} {\rm{d}}x}} = {e^{\frac{1}{2}\ln (1 – {x^2})}}\) A1
\( = \sqrt {1 – {x^2}} \) A1
multiply by \(I\) and attempt to integrate (M1)
\(y{(1 – {x^2})^{\frac{1}{2}}} = \int {\frac{1}{{\sqrt {1 – {x^2}} }}} {\rm{d}}x\) (A1)
\(y{(1 – {x^2})^{\frac{1}{2}}} = \arcsin x + c\) A1
[7 marks]
(i) attempt to find c M1
\(\frac{\pi }{2} = 0 + c\) A1
so \(f(x) = \frac{{\arcsin x + \frac{\pi }{2}}}{{\sqrt {1 – {x^2}} }}\) AG
(ii) \(\mathop {\lim }\limits_{x \to – 1} f(x) = \frac{0}{0}\) , so attempt l’Hôpital’s rule (M1)
consider \(\mathop {\lim }\limits_{x \to – 1} \frac{{\frac{1}{{{{(1 – {x^2})}^{\frac{1}{2}}}}}}}{{\frac{{ – x}}{{{{(1 – {x^2})}^{\frac{1}{2}}}}}}}\) A1A1
\( = 1\) A1
[6 marks]
Question
Find the general solution of the differential equation \((1 – {x^2})\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 1 + xy\) , for \(\left| x \right| < 1\) .
(i) Show that the solution \(y = f(x)\) that satisfies the condition \(f(0) = \frac{\pi }{2}\) is \(f(x) = \frac{{\arcsin x + \frac{\pi }{2}}}{{\sqrt {1 – {x^2}} }}\) .
(ii) Find \(\mathop {\lim }\limits_{x \to – 1} f(x)\) .
▶️Answer/Explanation
Markscheme
rewrite in linear form M1
\(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} + \left( {\frac{{ – x}}{{1 – {x^2}}}} \right)y = \frac{1}{{1 – {x^2}}}\)
attempt to find integrating factor M1
\(I = {e^{\int {\frac{{ – x}}{{1 – {x^2}}}} {\rm{d}}x}} = {e^{\frac{1}{2}\ln (1 – {x^2})}}\) A1
\( = \sqrt {1 – {x^2}} \) A1
multiply by \(I\) and attempt to integrate (M1)
\(y{(1 – {x^2})^{\frac{1}{2}}} = \int {\frac{1}{{\sqrt {1 – {x^2}} }}} {\rm{d}}x\) (A1)
\(y{(1 – {x^2})^{\frac{1}{2}}} = \arcsin x + c\) A1
[7 marks]
(i) attempt to find c M1
\(\frac{\pi }{2} = 0 + c\) A1
so \(f(x) = \frac{{\arcsin x + \frac{\pi }{2}}}{{\sqrt {1 – {x^2}} }}\) AG
(ii) \(\mathop {\lim }\limits_{x \to – 1} f(x) = \frac{0}{0}\) , so attempt l’Hôpital’s rule (M1)
consider \(\mathop {\lim }\limits_{x \to – 1} \frac{{\frac{1}{{{{(1 – {x^2})}^{\frac{1}{2}}}}}}}{{\frac{{ – x}}{{{{(1 – {x^2})}^{\frac{1}{2}}}}}}}\) A1A1
\( = 1\) A1
[6 marks]
Question
Use l’Hôpital’s rule to find \(\mathop {\lim }\limits_{x \to 0} (\csc x – \cot x)\).
▶️Answer/Explanation
Markscheme
\(\mathop {\lim }\limits_{x \to 0} (\csc x – \cot x) = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{1 – \cos x}}{{\sin x}}} \right)\) M1A1
\( = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}}{{\cos x}}} \right)\) M1A1
\( = 0\) A1
Question
Let
\({I_n} = \int_1^\infty {{x^n}{{\text{e}}^{ – x}}{\text{d}}x} \) where \(n \in \mathbb{N}\).
Using l’Hôpital’s rule, show that
\(\mathop {\lim }\limits_{x \to \infty } {x^n}{{\text{e}}^{ – x}} = 0\) where \(n \in \mathbb{N}\).
Show that, for \(n \in {\mathbb{Z}^ + }\),
\[{I_n} = \alpha {{\text{e}}^{ – 1}} + \beta n{I_{n – 1}}\]
where \(\alpha \), \(\beta \) are constants to be determined.
Determine the value of \({I_3}\), giving your answer as a multiple of \({{\text{e}}^{ – 1}}\).
▶️Answer/Explanation
Markscheme
consider \(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^x}}}\) M1
its value is \(\frac{\infty }{\infty }\) so we use l’Hôpital’s rule
\(\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{{\text{e}}^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{n{x^{n – 1}}}}{{{{\text{e}}^x}}}\) (A1)
its value is still \(\frac{\infty }{\infty }\) so we need to differentiate numerator and denominator a further \(n – 1\) times (R1)
this gives \(\mathop {\lim }\limits_{x \to \infty } \frac{{n!}}{{{{\text{e}}^x}}}\) A1
since the numerator is finite and the denominator \( \to \infty \), the limit is zero AG
[4 marks]
attempt at integration by parts \(\left( {{I_n} = – \int_1^\infty {{x^n}{\text{d}}({{\text{e}}^{ – x}})} } \right)\) M1
\({I_n} = – [{x^n}{{\text{e}}^{ – x}}]_1^\infty + n\int_1^\infty {{x^{n – 1}}{{\text{e}}^{ – x}}{\text{d}}x} \) A1A1
\( = {{\text{e}}^{ – 1}} + n{I_{n – 1}}\) A1
\(\alpha = \beta = 1\)
[9 marks]
\({I_3} = {{\text{e}}^{ – 1}} + 3{I_2}\) M1
\( = {{\text{e}}^{ – 1}} + 3({{\text{e}}^{ – 1}} + 2{I_1})\) A1
\( = 4{{\text{e}}^{ – 1}} + 6({{\text{e}}^{ – 1}} + {I_0})\) A1
\( = 4{{\text{e}}^{ – 1}} + 6{{\text{e}}^{ – 1}} + 6\int_1^\infty {{{\text{e}}^{ – x}}{\text{d}}x} \)
\( = 10{{\text{e}}^{ – 1}} – 6[{{\text{e}}^{ – x}}]_1^\infty \) A1
\( = 16{{\text{e}}^{ – 1}}\) A1
[9 marks]
Question
QUESTION 1
QUESTION 1
Using l’Hôpital’s Rule, determine the value of\[\mathop {\lim }\limits_{x \to 0} \frac{{\tan x – x}}{{1 – \cos x}} .\]
▶️Answer/Explanation
Markscheme
MARKSCHEME 1
MARKSCHEME 1
\(\mathop {\lim }\limits_{x \to 0} \frac{{\tan x – x}}{{1 – \cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sec }^2}x – 1}}{{\sin x}}\) M1A1A1
this still gives \(\frac{0}{0}\)
EITHER
repeat the process M1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{2{{\sec }^2}x\tan x}}{{\cos x}}\) A1
\( = 0\) A1
OR
\( = \mathop {\lim }\limits_{x \to 0} \frac{{{{\tan }^2}x}}{{\sin x}}\) M1
\( = \mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{{{{\cos }^2}x}}\) A1
\( = 0\) A1
[6 marks]
Question
Find the limits
- \(\lim_{x\to 1} \frac{x^2+1}{(x-1)^2}\)
- \(\lim_{x\to 1^{+}} \frac{x^2-1}{(x-1)^2}\)
- \(\lim_{x\to 1^{-}} \frac{x^2-1}{(x-1)^2}\)
▶️Answer/Explanation
Ans:
- \(\lim_{x\to 1} \frac{x^2+1}{(x-1)^2}=+\infty\)
- \(\lim_{x\to 1^{+}} \frac{x^2-1}{(x-1)^2}=\lim_{x\to 1^{+}}\frac{x+1}{x-1}=+\infty\)
- \(\lim_{x\to 1^{-}} \frac{x^2-1}{(x-1)^2}=\lim_{x\to 1^{-}}\frac{x+1}{x-1}=-\infty\)
Question
Find the limit \(\lim_{x\to 0} \frac{2\cos x -2+5x^2}{x^2}\)
▶️Answer/Explanation
Ans:
\(\lim_{x\to 0} \frac{2\cos x -2+5x^2}{x^2} \overset{(\frac{0}{0})}{=}\lim_{x\to 0}\frac{-2\sin x+10x}{2x}=\lim_{x\to 0}(\frac{-\sin x}{x}+5)=-1+5=4\)
Question
By using l’Hopital, show that
- \(\lim_{x\to \pm\infty} \frac{ax+b}{dx+e}=\frac{a}{d}\)
- \(\lim_{x\to \pm\infty} \frac{ax^2+bx+c}{dx^2+ex+f}=\frac{a}{d}\)
- \(\lim_{x\to \pm\infty} \frac{ax+b}{dx^2+ex+f}=0\)
- What is the geometric interpretation of these results for the corresponding graphs?
▶️Answer/Explanation
Ans:
- \(\lim_{x\to \pm\infty} \frac{ax+b}{dx+e}\overset{(\frac{\infty}{\infty})}{=}\lim_{x\to \pm\infty}\frac{a}{d} =\frac{a}{d}\)
- \(\lim_{x\to \pm\infty} \frac{ax^2+bx+c}{dx^2+ex+f}\overset{(\frac{\infty}{\infty})}{=}\lim_{x\to \pm\infty}\frac{2ax+b}{2dx+e}\overset{\frac{\infty}{\infty}}{=} \lim_{x\to \pm\infty}\frac{2a}{2d} =\frac{a}{d}\)
- \(\lim_{x\to \pm\infty} \frac{ax+b}{dx^2+ex+f}\overset{(\frac{\infty}{\infty})}{=}\lim_{x\to \pm\infty}\frac{a}{2dx+e}=0\)
- They give the horizontal asymptote .
for (i) and (ii) the line \(y=\frac{a}{d},\) for (iii) the line \(y=0\)
Question
By using l’Hopital, show that
- \(\lim_{x\to \pm\infty} \frac{ax+b}{dx+e}=\frac{a}{d}\)
- \(\lim_{x\to \pm\infty} \frac{ax^2+bx+c}{dx^2+ex+f}=\frac{a}{d}\)
- \(\lim_{x\to \pm\infty} \frac{ax+b}{dx^2+ex+f}=0\)
- What is the geometric interpretation of these results for the corresponding graphs?
▶️Answer/Explanation
Ans:
- \(\lim_{x\to \pm\infty} \frac{ax+b}{dx+e}\overset{(\frac{\infty}{\infty})}{=}\lim_{x\to \pm\infty}\frac{a}{d} =\frac{a}{d}\)
- \(\lim_{x\to \pm\infty} \frac{ax^2+bx+c}{dx^2+ex+f}\overset{(\frac{\infty}{\infty})}{=}\lim_{x\to \pm\infty}\frac{2ax+b}{2dx+e}\overset{\frac{\infty}{\infty}}{=} \lim_{x\to \pm\infty}\frac{2a}{2d} =\frac{a}{d}\)
- \(\lim_{x\to \pm\infty} \frac{ax+b}{dx^2+ex+f}\overset{(\frac{\infty}{\infty})}{=}\lim_{x\to \pm\infty}\frac{a}{2dx+e}=0\)
- They give the horizontal asymptote .
for (i) and (ii) the line \(y=\frac{a}{d},\) for (iii) the line \(y=0\)
Question
Find the limits
- \(\lim_{x\to +\infty} \frac{3x^2+x+e^x}{x^2+5x+3}\)
- \(\lim_{x\to -\infty} \frac{3x^2+x+e^x}{x^2+5x+3}\)
▶️Answer/Explanation
Ans:
- \(\lim_{x\to +\infty} \frac{3x^2+x+e^x}{x^2+5x+3}\overset{(\frac{\infty}{\infty})}{=} \lim_{x\to +\infty}\frac{6x+1+e^x}{2x+5}\overset{(\frac{\infty}{\infty})}{=}\lim_{x\to +\infty}\frac{6+e^x}{2}=+\infty \)
- \(\lim_{x\to -\infty} \frac{3x^2+x+e^x}{x^2+5x+3}\overset{(\frac{\infty}{\infty})}{=} \lim_{x\to -\infty}\frac{6x+1+e^x}{2x+5}\overset{(\frac{\infty}{\infty})}{=}\lim_{x\to -\infty}\frac{6+e^x}{2}=3 \)(since \(e^x \to 0\))
Question
Find the limits
- \(\lim_{x\to 0}\frac{e^{2x}-1}{2x}\)
- \(\lim_{x\to 0}\frac{e^{2x}-1-2x}{x^2}\)
▶️Answer/Explanation
Ans:
- \(\lim_{x\to 0}\frac{e^{2x}-1}{2x}\overset{(\frac{0}{0})}{=}\lim_{x\to 0} \frac{2e^{2x}}{2} =1 \)
- \(\lim_{x\to 0}\frac{e^{2x}-1-2x}{x^2}\overset{(\frac{0}{0})}{=}\lim_{x\to 0}\frac{2e^{2x}-2}{2x}= \lim_{x\to 0}\frac{e^{2x}-1}{x}\overset{(\frac{0}{0})}{=}\lim_{x\to 0}\frac{2e^{2x}}{1}=2 \)
Question
Find the limits
- \(\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{x})\)
- \(\lim_{x\to 0^{+}}(\frac{1}{x+1}-\frac{1}{x})\)
- \(\lim_{x\to 0^{-}}(\frac{1}{x+1}-\frac{1}{x})\)
▶️Answer/Explanation
Ans:
- \(\lim_{x\to 0}(\frac{1}{x^2}-\frac{1}{x})=\lim_{x\to 0}(\frac{1-x}{x^2})=+\infty \)
- \(\lim_{x\to 0^{+}}(\frac{1}{x+1}-\frac{1}{x})=\lim_{x\to 0^{+}}\frac{x-x-1}{(x+1)x}=\lim_{x\to 0^{+}}\frac{-1}{(x+1)x}=-\infty\)
- \(\lim_{x\to 0^{-}}(\frac{1}{x+1}-\frac{1}{x})=\lim_{x\to 0^{-}}\frac{x-x-1}{(x+1)x}=\lim_{x\to 0^{-}}\frac{-1}{(x+1)x}=+\infty\)
Question
Find the limits
- \(\lim_{x\to 1}\frac{3\ln x}{x-1}\)
- \(\lim_{x\to 1}\frac{x-1}{3\ln x}\)
▶️Answer/Explanation
Ans:
- \(\lim_{x\to 1}\frac{3\ln x}{x-1}\overset{(\frac{0}{0})}{=}\lim_{x\to 1}\frac{\frac{3}{x}}{1}=3\)
- \(\lim_{x\to 1}\frac{x-1}{3\ln x}\overset{(\frac{0}{0})}{=}\lim_{x\to 1}\frac{1}{\frac{3}{x}}=\frac{1}{3}\)
Question
- Find the limit \(\lim_{x\to 0}\frac{\ln (1+ax)}{x}\)
- Hence find \(\lim_{n\to +\infty}n \ln (1+\frac{a}{n})\)
- Hence show that \(\lim_{n\to +\infty}n \ln (1+\frac{a}{n})^n=e^a\)
- Given that \(FV=PV(1+\frac{r}{100k})^{nk},\) find \(\lim_{k\to +\infty}FV\)
▶️Answer/Explanation
Ans:
- \(\lim_{x\to 0}\frac{\ln (1+ax)}{x}\overset{(\frac{0}{0})}{=}\lim_{x\to 0}\frac{\frac{a}{1+ax}}{1}=a\)
- \(\lim_{n\to +\infty}n \ln (1+\frac{a}{n})=\lim_{n\to +\infty}\frac{\ln (1+\frac{a}{n})}{\frac{1}{n}}=a\)(set \(x=\frac{1}{n}\))
- \(\lim_{n\to +\infty}(1+\frac{a}{n})^n=\lim_{n\to +\infty}e^{\ln (1+\frac{a}{n})^n} =\lim_{n\to +\infty}e^{n \ln (1+\frac{a}{n})}=e^a \)
- \(\lim_{k\to +\infty}FV= \lim_{k\to +\infty} PV \left [ (1+\frac{\frac{r}{100}}{k})^k \right ] ^n=PV(e^{\frac{r}{100}})^n=PVe^{\frac{rn}{100}}\)
Question
- Find the limit \(\lim_{x\to 0}\frac{\ln (1+ax)}{x}\)
- Hence find \(\lim_{n\to +\infty}n \ln (1+\frac{a}{n})\)
- Hence show that \(\lim_{n\to +\infty}n \ln (1+\frac{a}{n})^n=e^a\)
- Given that \(FV=PV(1+\frac{r}{100k})^{nk},\) find \(\lim_{k\to +\infty}FV\)
▶️Answer/Explanation
Ans:
- \(\lim_{x\to 0}\frac{\ln (1+ax)}{x}\overset{(\frac{0}{0})}{=}\lim_{x\to 0}\frac{\frac{a}{1+ax}}{1}=a\)
- \(\lim_{n\to +\infty}n \ln (1+\frac{a}{n})=\lim_{n\to +\infty}\frac{\ln (1+\frac{a}{n})}{\frac{1}{n}}=a\)(set \(x=\frac{1}{n}\))
- \(\lim_{n\to +\infty}(1+\frac{a}{n})^n=\lim_{n\to +\infty}e^{\ln (1+\frac{a}{n})^n} =\lim_{n\to +\infty}e^{n \ln (1+\frac{a}{n})}=e^a \)
- \(\lim_{k\to +\infty}FV= \lim_{k\to +\infty} PV \left [ (1+\frac{\frac{r}{100}}{k})^k \right ] ^n=PV(e^{\frac{r}{100}})^n=PVe^{\frac{rn}{100}}\)
Question
By using the fact \(\lim_{x\to 0}\frac{\sin x}{x}=1\) (without using l’Hopital’s rule), find the following limits
| a. \(\lim_{x\to 0} \frac{\sin 3x}{3x}\) | |
| b. \(\lim_{x\to 0} \frac{\sin 2x}{x}\) | |
| c. \(\lim_{x\to 0} \frac{\sin 5x}{3x}\) | |
| d. \(\lim_{x\to 2} \frac{\sin (x-2)}{x-2}\) | |
| e. \(\lim_{x\to 0} \frac{3x+2 \sin x}{x}\) | |
| f. \(\lim_{x\to 0^{+}} \frac{\sin x}{x^2}\) | |
| g. \(\lim_{x\to 0^{+}} \frac{\sin x}{\sqrt{x}}\) | |
| h. \(\lim_{x\to 0} \frac{\sin 2x}{x \cos x}\) | |
| i. \(\lim_{n\to +\infty} n \sin \frac{1}{n}\) | |
| j. \(\lim_{n\to +\infty} n^2 \sin \frac{1}{n}\) | |
| k. \(\lim_{n\to +\infty} n \sin \frac{1}{n^2}\) |
▶️ Answer/Explanation
Ans:
| a. \(\lim_{x\to 0} \frac{\sin 3x}{3x}\) | \(=1\) (let \(y=3x\), so \(y \to 0\) when \(x \to 0\)) |
| b. \(\lim_{x\to 0} \frac{\sin 2x}{x}\) | \(=2\lim_{x\to 0} \frac{\sin 2x}{2x}=2\) |
| c. \(\lim_{x\to 0} \frac{\sin 5x}{3x}\) | \(=\frac{5}{3}\lim_{x\to 0}\frac{\sin 5x}{5x}=\frac{5}{3}\) |
| d. \(\lim_{x\to 2} \frac{\sin (x-2)}{x-2}\) | \(=1\) (let \(y=x-2\), so \(y \to 0\) when \(x \to 2\) |
| e. \(\lim_{x\to 0} \frac{3x+2 \sin x}{x}\) | \(=\lim_{x\to 0}(3+2\frac{\sin x}{x})=3+2=5\) |
| f. \(\lim_{x\to 0^{+}} \frac{\sin x}{x^2}\) | \(=\lim_{x\to 0^{+}}\frac{1}{x}\frac{\sin x}{x}=+\infty (+\infty×1=+\infty)\) |
| g. \(\lim_{x\to 0^{+}} \frac{\sin x}{\sqrt{x}}\) | \(=\lim_{x\to 0^{+}}\sqrt{x}\frac{\sin x}{x}=0×1=0\) |
| h. \(\lim_{x\to 0} \frac{\sin 2x}{x \cos x}\) | \(=\lim_{x\to 0}\frac{2\sin x \cos x}{x \cos x}=\lim_{x\to 0}\frac{2 \sin x}{x}=2\) |
| i. \(\lim_{n\to +\infty} n \sin \frac{1}{n}\) | \(=\lim_{n\to +\infty}\frac{\sin \frac{1}{n}}{\frac{1}{n}} =1\)(let \(y=\frac{1}{n}\), so \(y \to 0\) when \(n \to +\infty\)) |
| j. \(\lim_{n\to +\infty} n^2 \sin \frac{1}{n}\) | \(=\lim_{n\to +\infty}n\frac{\sin \frac{1}{n}}{\frac{1}{n}} =+\infty (+\infty×1=+\infty)\) |
| k. \(\lim_{n\to +\infty} n \sin \frac{1}{n^2}\) | \(=\lim_{n\to +\infty}\frac{1}{n}\frac{\sin \frac{1}{n}}{\frac{1}{n^2}}=0×1=0\) |
Question
Let \(f(x)=\frac{2\tan x-\tan 2x}{\sin 2x-2 \sin x}\)
- The function is not defined at \(x = 0\). Look at the graph of \(f\) in the GDC and find values of \(f (x)\) near \(x = 0\) and thus guess the value of \(\lim_{x\to 0} f(x)\)
- Confirm your guess by using l’Hôpital’s rule.
▶️Answer/Explanation
Ans:
- It takes values close to 2, so the limit seems to be 2.
- \(\lim_{x\to 0} \frac{2\tan x-\tan 2x}{\sin 2x-2 \sin x}\overset{(\frac{0}{0})}{=}\lim_{x\to 0}\frac{2\sec^2 x-2\sec^2 2x}{2 \cos 2x-2 \cos x}=\lim_{x\to 0}\frac{\sec^2 x-\sec^2 2x}{\cos 2x-\cos x}\)
\(=\lim_{x\to 0}\frac{\frac{1}{\cos^2 x}-\frac{1}{\cos^2 2x}}{\cos 2x-\cos x}=\lim_{x\to 0}\frac{\frac{\cos^2 2x-\cos^2 x}{\cos^2 x \cos^2 2x}}{\cos 2x-\cos x}=\lim_{x\to 0}\frac{\cos 2x+\cos x}{\cos^2 x \cos^2 2x}=2\)
Question
Hence, or otherwise, determine the value of \(\lim_{x\to 0} \frac{2 \ln (1+e^x)-x-\ln 4}{x^2}\).
▶️Answer/Explanation
Ans:
using l’Hôpital’s Rule
\(\lim_{x\to 0} \frac{2 \ln (1+e^x)-x-\ln 4}{x^2}=\lim_{x\to 0} \frac{2e^x\div(1+e^x)-1}{2x}\)
=\(\lim_{x\to 0} \frac{2e^x\div(1+e^x)^2}{2}=\frac{1}{4}\)
Question
Find the value of \(\lim_{x\to 0}(\frac{1}{x}-\cot x)\)
▶️Answer/Explanation
Ans:
EITHER
\(\lim_{x\to 0}(\frac{1}{x}-\cot x)=\lim_{x\to 0}(\frac{\tan x-x}{x \tan x})\)
\(=\lim_{x\to 0}(\frac{\sec^2 x-1}{x \sec^2x+\tan x})\), using l’Hopital
\(=\lim_{x\to 0}(\frac{2\sec^2 x \tan x}{2\sec^2 x+2x \sec^2 x \tan x})\)
\(=0\)
OR
\(\lim_{x\to 0}(\frac{1}{x}-\cot x)=\lim_{x\to 0}(\frac{\sin x – x \cos x}{x \sin x})\)
\(=\lim_{x\to 0}(\frac{x\sin x}{\sin x+x \cos x}),\) using l’Hopital
\(=\lim_{x\to 0}(\frac{\sin x+x \cos x}{2 \cos x-x \sin x})\)
\(=0\)
Question
Find
- \(\lim_{x\to 0} \frac{\tan x}{x+x^2}\);
- \(\lim_{x\to 1} \frac{1-x^2+2x^2 \ln x}{1-\sin \frac{\pi x}{2}}\)
▶️Answer/Explanation
Ans:
- \(\lim_{x\to 0} \frac{\tan x}{x+x^2}=\lim_{x\to 0}\frac{\sec^2 x}{1+2x}\)
\(\lim_{x\to 0} \frac{\tan x}{x+x^2}=\frac{1}{1}=1\) - \(\lim_{x\to 1} \frac{1-x^2+2x^2 \ln x}{1-\sin \frac{\pi x}{2}}=\lim_{x\to 1} \frac{-2x+2x+4x \ln x}{-\frac{\pi}{2}\cos \frac{\pi x}{2}}\)
\(\lim_{x\to 1}\frac{4+4 \ln x}{\frac{\pi^2}{4}\sin \frac{\pi x}{2}}\)
\(\lim_{x\to 1}\frac{1-x^2+2x^2 \ln x}{1-\sin \frac{\pi x}{2}}=\frac{4}{\frac{\pi^2}{4}}=\frac{16}{\pi^2}\)
Question
Find
- \(\lim_{x\to 0} \frac{\tan x}{x+x^2}\);
- \(\lim_{x\to 1} \frac{1-x^2+2x^2 \ln x}{1-\sin \frac{\pi x}{2}}\)
▶️Answer/Explanation
Ans:
- \(\lim_{x\to 0} \frac{\tan x}{x+x^2}=\lim_{x\to 0}\frac{\sec^2 x}{1+2x}\)
\(\lim_{x\to 0} \frac{\tan x}{x+x^2}=\frac{1}{1}=1\) - \(\lim_{x\to 1} \frac{1-x^2+2x^2 \ln x}{1-\sin \frac{\pi x}{2}}=\lim_{x\to 1} \frac{-2x+2x+4x \ln x}{-\frac{\pi}{2}\cos \frac{\pi x}{2}}\)
\(\lim_{x\to 1}\frac{4+4 \ln x}{\frac{\pi^2}{4}\sin \frac{\pi x}{2}}\)
\(\lim_{x\to 1}\frac{1-x^2+2x^2 \ln x}{1-\sin \frac{\pi x}{2}}=\frac{4}{\frac{\pi^2}{4}}=\frac{16}{\pi^2}\)
Question
Using l’Hopital’s Rule determine the value of \(\lim_{x\to 0}\frac{\tan x-x}{1-\cos x}\)
▶️Answer/Explanation
Ans:
\(\lim_{x\to 0}\frac{\tan x-x}{1-\cos x}\overset{(\frac{0}{0})}{=}\lim_{x\to 0}\frac{\sec^2 x-1}{\sin x}\overset{(\frac{0}{0})}{=}\lim_{x\to 0}\frac{2\sec x \sec x \tan x}{\cos x}=0\)
Question
Use l’Hopital’s Rule to find \(\lim_{x\to 0}(\csc x-\cot x)\)
▶️Answer/Explanation
Ans:
\(\lim_{x\to 0}(\frac{1}{\sin x}-\frac{\cos x}{\sin x})=\lim_{x\to 0}\frac{1-\cos x}{\sin x}\overset{(\frac{0}{0})}{=}\lim_{x\to 0}\frac{\sin x}{\cos x}=0\)
Question
Calculate
- \(\lim_{x\to 0}\frac{2^x-1}{x}\)
- \(\lim_{x\to 0}\frac{(1+x^2)^\frac{3}{2}-1}{\ln (1+x)-x}\)
▶️Answer/Explanation
Ans:
\(\lim_{x\to 0}\frac{2^x-1}{x}\overset{(\frac{0}{0})}{=}\lim_{x\to 0}\frac{2^x \ln 2}{1}=\ln 2\)
\(\lim_{x\to 0}\frac{(1+x^2)^\frac{3}{2}-1}{\ln (1+x)-x}\overset{(\frac{0}{0})}{=} \lim_{x\to 0}\frac{\frac{3}{2}(1+x^2)^\frac{1}{2}2x}{\frac{1}{1+x}-1}=\lim_{x\to 0}\frac{\frac{3}{2}(1+x^2)^\frac{1}{2}2x}{\frac{-x}{1+x}}=\lim_{x\to 0}[-3(1+x^2)^\frac{1}{2}(1+x)]=-3\)
Question
[theoretical]
- Write down the term in \(x^r\) in the expansion of \((x+h)^n\), where \(0 \le r \le n, n \in Z^{+}\)
- Hence differentiate \(x^n,n \in Z^{+}\), from first principles.
- Starting from the result \(x^n×x^{-n}=1,\) deduce the derivative of \(x^{-n},n \in Z^{+}\)
▶️Answer/Explanation
Ans:
- \(r^{th}\)term = \(\begin{pmatrix} n \\ n-r \end{pmatrix}x^rh^{n-r}\left (=\frac{n!}{r!(n-r)!}x^rh^{n-r} \right )\)
- \(\frac{d(x^n)}{dx}=\lim_{h\to 0} \left ( \frac{(x+h)^n-x^n}{h} \right )\)
\(=\lim_{h\to 0} \left ( \frac{x^n+\begin{pmatrix} n \\ 1 \end{pmatrix}x^{n-1}h+\begin{pmatrix} n \\ 2 \end{pmatrix}x^{n-2}h^2+…+h^n-x^n}{h} \right )\)
\(=\lim_{h\to 0}\left ( \frac{x^n+nx^{n-1}h+\frac{n(n-1)}{2}x^{n-2}h^2+…+h^n-x^n}{h} \right )\)
\(=\lim_{h\to 0}\left ( nx^{n-1}+\frac{n(n-1)}{2}x^{n-2}h+…+h^{n-1} \right )\)
\(=nx^{n-1}\) - \(x^n×x^{-n}=1\)
\(x^n\frac{d(x^{-n})}{dx}+x^{-n}\frac{d(x^n)}{dx}=0\)
\(x^n\frac{d(x^{-n})}{dx}+x^{-n}×nx^{n-1}=0\)
\(x^n\frac{d(x^{-n})}{dx}+nx^{-1}=0\)
\(\frac{d(x^{-n})}{dx}=\frac{-nx^{-1}}{x^n}(=-nx^{-(1+n)})\)