Question
Use l’Hôpital’s rule to find $\lim_{x\to 0} \frac{\sec^4 x – \cos^2 x}{x^4 – x^2}$.
▶️Answer/Explanation
Solution:-
$\lim_{x\to 0} \frac{4\sec^4 x \tan x + 2\sin x \cos x}{4x^3 – 2x}$
$= \lim_{x\to 0} \frac{16\sec^4 x \tan^2 x + 4\sec^6 x – 2\sin^2 x + 2\cos^2 x}{12x^2 – 2}$
$= 3$
Solution
$\lim_{x\to 0} \frac{4\sec^4 x \tan x + 2\sin x \cos x}{4x^3 – 2x}$, on putting x=0, it is still indeterminate form, therefore again appying L’Hospital rule we get,
$= \lim_{x\to 0} \frac{16\sec^4 x \tan^2 x + 4\sec^6 x – 2\sin^2 x + 2\cos^2 x}{12x^2 – 2}$, on putting x=0, we get
$= \frac{-3}{{2}$
Question
Consider the polynomial \(P(x)=3x^{3}+5x^{2}+x-1\)
(a) Show that (x+1) is a factor or P(x).
(b) Hence, express P(x) as a product of three linear factors.
Now consider the polynomial Q(x)=(x+1)(2x+1).
(c)Express \(\frac{1}{Q(x)}\) in the form \(\frac{A}{x+1}+\frac{B}{2x+1}\), where \(A, B \in \mathbb{Z}\).
(d) Hence, or otherwise, show that \(\frac{1}{(x+1)Q(x)}=\frac{4}{2x+1}-\frac{2}{x+1}-\frac{1}{(x+1)^{2}}\).
(e)Hence, find \(\int \frac{1}{(x+1)^{2}(2x+1)}dx\).
Consider the function defined by \(f(x)=\frac{P(x)}{(x+1)Q(x)}\); where \(x\neq -1, x\neq -\frac{1}{2}\).
(f) Find
(i) \(\displaystyle \lim_{x \to -1}f(x)\);
(ii) \(\displaystyle \lim_{x \to \infty }f(x)\).
▶️Answer/Explanation
Ans:
(a) Substituting x = -1 in the polynomial P(x) as given below, or applying the synthetic division (Horner’s Method) method taking the coefficients of polynomial, 3, 5, 1, -1 and the value of factor x = -1, or by long division, we get that (x+1) is a factor or P(x)
(b) using long division or synthetic division to divide P(x) by (x+1) we get from previous step,
\(P(x)=(x+1)(3x^{2}+2x-1)\)
\(=(x+1)(x+1)(3x-1)(=(x+1)^{2}(3x-1))\)
(c)According to the question, \(\frac{1}{Q(x)}\) =
substitute the values of x = \(-1\) and \(-\frac{1}{2}\), we get
\(1=-A\) and \(1=\frac{1}{2}B\)
Therefore, \(A=-1\) and \(B=2\)
Therefore, \(\frac{1}{(x+1)(2x+1)}=-\frac{1}{x+1}+\frac{2}{2x+1}\)
(d) Taking the result obtained in (c) forward, according to the question
(e) Since, \(\frac{1}{(x+1)^{2}(2x+1)}\) = 
Therefore,
(f) (i) METHOD 1
By cancelling the factors in numerator and denominator and then substitute \(x=-1\)
METHOD 2
By expanding denominator, then differentiating numerator and denominator twice and substitute \(x=-1\)
(ii) METHOD 1
divide all terms by \(x^{3}\)
METHOD 2
By cancelling the factors and considering coefficients of x or by dividing all terms by x
METHOD 3
By expanding denominator, then differentiating numerator and denominator three times
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:
