Question

The value of \[\mathop {\lim }\limits_{x \to 1} \frac{{\sin ({e^{x - 1}} - 1)}}{{\ln x}}\]=
A) 0
B) e
C) \[\frac{1}{e}\]
D) 1

Answer 1

Hint: This question can be solved by two methods. The first method is that we will simplify the expression and then put the limit. The second method is that we will use the L- hospital rule to determine the suitable answer. In this question, we will use the L- hospital rule. L – hospital rule is applicable only when the \[\frac{0}{0}\] or \[\frac{\infty }{\infty }\] form arises.

Formula Used:
 L- hospital rule: L – hospital rule is applicable only when the \[\frac{0}{0}\] or \[\frac{\infty }{\infty }\] form arises.
\[\cos 0^0\,=1\]

Complete step by step Solution:
First of all, put the limit into the function to check the \[\frac{0}{0}\]or \[\frac{\infty }{\infty }\]form. If the form arises, then apply the L -hospital rule,
Let us assume that there are two function f(x) and g(x) in the form of fraction with the limit such as \[\mathop {\lim }\limits_{x \to a} \frac{{f(x)}}{{g(x)}}\]. If we take the limit and get the intermediate form such as \[\frac{0}{0}\]or \[\frac{\infty }{\infty }\], then we will apply the l- hospital rule. For that, we will determine the differentiation of f(x) and g(x) respectively. And then we will take the limit. Therefore, we can write it as,
\[ \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{{f^{'}}(x)}}{{{g^{'}}(x)}}\]
Now, we have given that
\[ \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{\sin ({e^{x - 1}} - 1)}}{{\ln x}}\]
Here \[\frac{0}{0}\] arises therefore, we will apply the L – hospital rule. For that purpose, first, we will differentiate the numerator and after that, we will differentiate the denominator. Therefore, we will get,

\[ \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{\cos ({e^{x - 1}} - 1) \times {e^{x - 1}}}}{{\frac{1}{x}}}\]
Now, take the limit
\[ \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{\cos ({e^{1 - 1}} - 1) \times {e^{1 - 1}}}}{{\frac{1}{1}}}\]
And
\[ \Rightarrow \mathop {\lim }\limits_{x \to 1} \frac{{\cos 0 \times 1}}{{\frac{1}{1}}}\]
\[ \Rightarrow 1\]
Now, the final answer is 1.

Hence, the correct option is (D).

Note:It is important to note that the L- hospital rule is applicable only when the intermediate such as \[\frac{0}{0}\]or \[\frac{\infty }{\infty }\] arise. Therefore, check the intermediate form of the function by taking the limit first.