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

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

Explanation

Solution

First check if the given limit is in the form (0)0{{\left( 0 \right)}^{0}} or not. For (0)0{{\left( 0 \right)}^{0}} form simplify using the formula lim⁡x→0(f(x))g(x)=elim⁡x→0g(x)ln⁡(f(x))\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 (sin⁡0)0=(0)0{{\left( \sin 0 \right)}^{0}}={{\left( 0 \right)}^{0}}
The expression of the form lim⁡x→0(f(x))g(x)\displaystyle \lim_{x \to 0}{{\left( f\left( x \right) \right)}^{g\left( x \right)}} with the value (0)0{{\left( 0 \right)}^{0}} can be simplified by taking as elim⁡x→0g(x)ln⁡(f(x)){{e}^{\displaystyle \lim_{x \to 0}g\left( x \right)\ln \left( f\left( x \right) \right)}}
Considering our equation lim⁡x→0+(sin⁡x)x\displaystyle \lim_{x \to {{0}^{+}}}{{\left( \sin x \right)}^{x}}
By comparison, f(x)=sin⁡xf\left( x \right)=\sin x and g(x)=xg\left( x \right)=x
So, it can be simplified as
⇒elim⁡x→0+xln⁡(sin⁡x)\Rightarrow {{e}^{\displaystyle \lim_{x \to {{0}^{+}}}x\ln \left( \sin x \right)}}
Taking the value of ‘x’ as ‘0’, we get
xln⁡(sin⁡x)=0×ln⁡(sin⁡x)=0x\ln \left( \sin x \right)=0\times \ln \left( \sin x \right)=0
Putting this value in the equation, we get

& \Rightarrow {{e}^{0}} \\\ & \Rightarrow 1 \\\ \end{aligned}$$ 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.