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# How do you find $\displaystyle \lim_{x \to {{0}^{+}}}{{\left( \sin x \right)}^{x}}$?

Last updated date: 11th Aug 2024
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Hint: First check if the given limit is in the form ${{\left( 0 \right)}^{0}}$ or not. For ${{\left( 0 \right)}^{0}}$ form simplify using the formula $\displaystyle \lim_{x \to 0}{{\left( f\left( x \right) \right)}^{g\left( x \right)}}={{e}^{\displaystyle \lim_{x \to 0}g\left( x \right)\ln \left( f\left( x \right) \right)}}$. After simplification, put the value of ‘x’ as ‘0’ and do the necessary calculations to get the limiting value.

Complete step-by-step solution:
Putting the value of ‘x’ in the function we are getting ${{\left( \sin 0 \right)}^{0}}={{\left( 0 \right)}^{0}}$
The expression of the form $\displaystyle \lim_{x \to 0}{{\left( f\left( x \right) \right)}^{g\left( x \right)}}$ with the value ${{\left( 0 \right)}^{0}}$ can be simplified by taking as ${{e}^{\displaystyle \lim_{x \to 0}g\left( x \right)\ln \left( f\left( x \right) \right)}}$
Considering our equation $\displaystyle \lim_{x \to {{0}^{+}}}{{\left( \sin x \right)}^{x}}$
By comparison, $f\left( x \right)=\sin x$ and $g\left( x \right)=x$
So, it can be simplified as
$\Rightarrow {{e}^{\displaystyle \lim_{x \to {{0}^{+}}}x\ln \left( \sin x \right)}}$
Taking the value of ‘x’ as ‘0’, we get
$x\ln \left( \sin x \right)=0\times \ln \left( \sin x \right)=0$
Putting this value in the equation, we get
\begin{align} & \Rightarrow {{e}^{0}} \\ & \Rightarrow 1 \\ \end{align}
Hence, $\displaystyle \lim_{x \to {{0}^{+}}}{{\left( \sin x \right)}^{x}}=1$
Where, ‘1’ is the limiting value of the given limit.
This is the required solution of the given question.

Note: We know in logarithmic function the value of log which we are taking must be greater than ‘0’. So, in $\ln \left( \sin x \right)$, the value of $\sin x$ must be greater than ‘0’. Again as we know the range of $\sin x$ is $\left[ -1,1 \right]$, but since we have to get only those values which are greater than ‘0’, so now the range becomes $\left( 0,1 \right]$. Hence, multiplying this range of $\ln \left( \sin x \right)$ with ‘0’ we are getting $x\ln \left( \sin x \right)=0\times \ln \left( \sin x \right)=0$ as the value of $\sin x$ with range $\left( 0,1 \right]$ will always be a positive value.