Question

Find the limit of ${x^{\sin \left( x \right)}}$ as $x$ approaches 0.

Answer 1

We know that we say the limit of $f\left( x \right)$ is $L$ as $x$ approaches $a$, i.e. $\mathop {\lim }\limits_{x \to a} f\left( x \right) = L$
Also we have to make $f\left( x \right)$ as close to $L$ and $x$ as close to $a$ from both the sides and not letting it be $a$. So by using the above statement and expression we can solve the given question.

Complete step by step solution:
Given
${x^{\sin \left( x \right)}}............................\left( i \right)$

Now on comparing the question and the statement of limit we can say that$a = 0\,\,{\text{and}}\,\,{\text{f}}\left( x \right) = {x^{\sin x}}$. Since here ${\text{f}}\left( x \right) = {x^{\sin x}}$ it’s difficult to find its limit directly so it’s better to use some logarithmic functions since in some logarithmic functions the power can be extracted and used as a multiplier.

So converting our given ${\text{f}}\left( x \right) = {x^{\sin x}}$ to a form where we can directly find its limit.

Let:
$L = \mathop {\lim }\limits_{x \to 0} {x^{\sin x}}........................(ii)$

Now let’s take $\ln$ on both sides since we can use the power as a multiplier by the power rule.
$\Rightarrow \ln L = \ln \mathop {\lim }\limits_{x \to 0} {x^{\sin x}} \\\ \Rightarrow \ln L = \mathop {\lim }\limits_{x \to 0} \ln {x^{\sin x}} \\\$
Now by using the power rule we can write:
$\Rightarrow \ln L = \mathop {\lim }\limits_{x \to 0} \sin x\ln x \\\ \Rightarrow \ln L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln x}}{{\left( {\dfrac{1}{{\sin x}}} \right)}} \\\ \Rightarrow \ln L = \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln x}}{{{\text{cosec}}x}}....................\left( {iii} \right) \\\$
Now if we apply the limit zero to both numerator and denominator we see that we get $\dfrac{\infty }{\infty }$ which is indeterminate.

So in such cases we have to apply L’Hospital’s Rule.

It simply states that whenever we get an indeterminate form we have to find the derivative of both numerator and denominator of the fraction and apply the limit.

So applying L’Hospital’s Rule in (iii), we get:
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\ln x}}{{{\text{cosec}}x}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\left( {\dfrac{1}{x}} \right)}}{{ - \cos {\text{ec}}x\cot x}} \\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \mathop { - \lim }\limits_{x \to 0} \dfrac{{\sin x\tan x}}{x}...............\left( {iv} \right) \\\$
Again if we apply the limit zero to both numerator and denominator we see that we get $\dfrac{\infty }{\infty }$which is indeterminate.

So again applying L’Hospital’s Rule in (iv), we get:
$\Rightarrow \mathop { - \lim }\limits_{x \to 0} \dfrac{{\sin x\tan x}}{x} = - \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos x\tan x + \sin x{{\sec }^{2x}}}}{1} \\\ \Rightarrow \mathop { - \lim }\limits_{x \to 0} \dfrac{{\sin x\tan x}}{x} = - 0..........................\left( v \right) \\\ \\\$
Now we can write from (iii):
$\Rightarrow \ln L = - 0 \\\ \Rightarrow L = {e^{ - 0}} = 1..........................\left( {vi} \right) \\\$
Therefore$\mathop {\lim }\limits_{x \to 0} {x^{\sin x}} = 1$.

Note: We know that:

$$\dfrac{{d\sin x\tan x}}{{dx}} = \sin x + \tan x\sec x \\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\sec ^2}x\sin x + \tan x\sec x \\\$$

The above given approach for finding limits for similar problems is preferred due to its easier steps and processes.