Question

If $f\left( x \right) = \dfrac{{\sin \left( {{e^{x - 2}} - 1} \right)}}{{\ln \left( {x - 1} \right)}}$, then $\mathop {\lim }\limits_{x \to 2} f\left( x \right)$ is equal to
A) $- 2$
B) $- 1$
C) 0
D) 1

Answer 1

Here, we will first apply L’Hopital’s rule to reduce the limit. When the limit has been simplified to an extent such that if we substitute the limit value then our denominator does not turn out to be 0, then, we will substitute the limiting value and remove the limit to find the required value of $\mathop {\lim }\limits_{x \to 2} f\left( x \right)$ which satisfies this limit.
Formula Used:
We will use the following formulas:

1. $\dfrac{{dy}}{{dx}}\sin x = \cos x$
2. $\dfrac{{dy}}{{dx}}{e^x} = {e^x}$
3. $\dfrac{{dy}}{{dx}}\log x = \dfrac{1}{x}$
4. ${a^0} = 1$

Complete step by step solution:
The given function is $f\left( x \right) = \dfrac{{\sin \left( {{e^{x - 2}} - 1} \right)}}{{\ln \left( {x - 1} \right)}}$.
We have to find $\mathop {\lim }\limits_{x \to 2} f\left( x \right)$.
Therefore,
$\mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \dfrac{{\sin \left( {{e^{x - 2}} - 1} \right)}}{{\log \left( {x - 1} \right)}}$
Now, in order to solve this limit, we will apply the L’Hopital’s rule, in which if we have an indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$, we differentiate the numerator and the denominator and then take the limit.
Now, applying L’Hopital’s rule and using the formula $\dfrac{{dy}}{{dx}}\sin x = \cos x$, $\dfrac{{dy}}{{dx}}{e^x} = {e^x}$ and $\dfrac{{dy}}{{dx}}\log x = \dfrac{1}{x}$, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \dfrac{{\cos \left( {{e^{x - 2}} - 1} \right)\left( {{e^{x - 2}}} \right)}}{{\dfrac{1}{{\left( {x - 1} \right)}}}}$
Simplifying the expression, we get
$\Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \mathop {\lim }\limits_{x \to 2} \cos \left( {{e^{x - 2}} - 1} \right)\left( {{e^{x - 2}}} \right)\left( {x - 1} \right)$
Now, substituting $x = 2$ and hence, removing the limit, we get,
$\Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \cos \left( {{e^{2 - 2}} - 1} \right)\left( {{e^{2 - 2}}} \right)\left( {2 - 1} \right)$
$\Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \cos \left( {{e^0} - 1} \right)\left( {{e^0}} \right)\left( 1 \right)$
Now, using the identity ${e^0} = 1$, we get,
$\Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = \cos 0^\circ \left( 1 \right)\left( 1 \right)$
By trigonometric tables, we know that, $\cos 0^\circ = 1$
$\Rightarrow \mathop {\lim }\limits_{x \to 2} f\left( x \right) = 1$
Hence, $\mathop {\lim }\limits_{x \to 2} f\left( x \right)$ is equal to 1.

Therefore, option D is the correct answer.

Note:
In mathematics, a limit is a value that a function approaches as the input or the index approaches some value. Limits are an essential element of calculus and are used to define continuity, integrals and derivatives. The idea of limits was first developed by ‘Archimedes of Syracuse’ to measure curved figures and the volume of a sphere in the third century B.C.