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Question: If \(f(x) = \frac{\sin(e^{x - 2} - 1)}{\log(x - 1)},\) then \(\lim_{x \rightarrow 2}f(x)\) is given ...

If f(x)=sin⁡(ex−2−1)log⁡(x−1),f(x) = \frac{\sin(e^{x - 2} - 1)}{\log(x - 1)}, then lim⁡x→2f(x)\lim_{x \rightarrow 2}f(x) is given by

A

– 2

B

–1

C

0

D

1

Answer

1

Explanation

Solution

lim⁡x→2f(x)=lim⁡x→2sin⁡(ex−2−1)log⁡(t+1)=lim⁡t→0sin⁡(et−1)log⁡(t+1).\lim_{x \rightarrow 2}f(x) = \lim_{x \rightarrow 2}\frac{\sin(e^{x - 2} - 1)}{\log(t + 1)} = \lim_{t \rightarrow 0}\frac{\sin(e^{t} - 1)}{\log(t + 1)}.

(Putting x = 2 + t)

=lim⁡x→∞sin⁡(et−1)et−1.et−1t.tlog⁡(1+t)= \lim_{x \rightarrow \infty}\frac{\sin(e^{t} - 1)}{e^{t} - 1}.\frac{e^{t} - 1}{t}.\frac{t}{\log(1 + t)} =lim⁡t→0sin⁡(et−1)et−1(11!+t2!+.....)[1(1−12t+13t2−.....)]= \lim_{t \rightarrow 0}\frac{\sin(e^{t} - 1)}{e^{t} - 1}\left( \frac{1}{1!} + \frac{t}{2!} + ..... \right)\left\lbrack \frac{1}{\left( 1 - \frac{1}{2}t + \frac{1}{3}t^{2} - ..... \right)} \right\rbrack=1.1.1= 1

[∵\becauseAs t→0,et−1→0t \rightarrow 0,e^{t} - 1 \rightarrow 0, ∴sin⁡(et−1)(et−1)=1\therefore\frac{\sin(e^{t} - 1)}{(e^{t} - 1)} = 1]