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Question
Calculate the following limit
\(\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} – 1}}{x}\) .
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
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]