Question

Evaluate:$\underset{x\to 0}{\lim }x{\log }_e\left(\sin x\right)$

• A. $-1$
• B. ${\log }_{e}1$
• C. $1$
• D. None of these

Answer 1

The correct option is B

${\log }_{e}1$

Explanation for the correct option:

Finding the value of $\underset{x\to 0}{\lim }x{\log }_e\left(\sin x\right)$

Step 1: Using the L' hospitality rule

Consider the given Equation as

$$\begin{gathered} I=\underset{x\to 0}{\lim }x{\log }_e\left(\sin x\right) \\ \text{Putting the limit}x\to 0 \\ I=0{\log }_e\left(\sin 0\right) \\ I=0{\log }_e\left(0\right) \\ I=0.\infty \left[\text{indeterminent form}\right] \end{gathered}$$

Using L'Hospital Rule,

$\underset{x\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

So first we simplify the given Equation and then evaluate,

$I=\underset{x\to 0}{\lim }x{\log }_e\left(\sin x\right)$

Rewriting $x\text{as}\frac{1}{\left({\displaystyle \frac{1}{x}}\right)}$

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{{\log }_e\left(\sin x\right)}{{\displaystyle \frac{1}{x}}} \\ I=\frac{{\log }_e\left(\sin 0\right)}{{\displaystyle \frac{1}{0}}} \\ I=\frac{\infty }{\infty }\left[\text{indeterminent form}\right] \end{gathered}$$

Step 2: Again, Using the L' hospitality rule

$\underset{x\to 0}{\lim }\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to 0}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

$$\begin{gathered} I=\underset{x\to 0}{\lim }\frac{{\log }_e\left(\sin x\right)}{{\displaystyle \frac{1}{x}}} \\ I=\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{\cos x}{\sin x}}}{{\displaystyle -\frac{1}{x^2}}}\mathbf{[}\mathbf{\text{Where}}\mathbf{,}\frac{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{x}}{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{x}}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{t}\mathbf{}\mathbf{x}\mathbf{]} \\ I=\underset{x\to 0}{\lim }\frac{cotx}{{\displaystyle \frac{1}{-x^2}}}\mathbf{}[\text{Where,}cot\mathrm{x}=\frac{1}{tan\mathrm{x}}] \\ I=\underset{x\to 0}{\lim }\frac{-x}{{\displaystyle \frac{\tan x}{x}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\text{where,}}\underset{\mathbf{x}\mathbf{\to }\mathbf{0}}{\mathbf{l}\mathbf{i}\mathbf{m}}\frac{\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{x}}{\mathbf{x}}\mathbf{=}\mathbf{1}\mathbf{]} \\ I=\underset{x\to 0}{\lim }-x \\ I=0 \\ I={\log }_{e}1\left[\text{Where},lo{g}_{e}1=0\right] \end{gathered}$$

Hence, option (B) is the correct answer.