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Question: Find the value of \(\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{(\sin x)}^{\dfrac{1}{x}}}\)....

Find the value of lim⁡x→0+ (sin⁡x)1x\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{(\sin x)}^{\dfrac{1}{x}}}.

Explanation

Solution

Hint : First, apply the limits directly to find out whether the value obtained is indeterminate form. Then apply L’Hospital Rule if the value of limit is indeterminate form. Remember the known formula lim⁡x→0 sin⁡xx=1\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1.

Complete step by step solution :
First we will directly apply the limits to the given expression.
lim⁡x→0+ (sin⁡x)1x=(sin⁡0)10\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{(\sin x)}^{\dfrac{1}{x}}}={{(\sin 0)}^{\dfrac{1}{0}}}
We know sin⁡0=0,10=∞\sin 0=0,\dfrac{1}{0}=\infty , so above equation becomes,
lim⁡x→0+ (sin⁡x)1x=0∞\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{(\sin x)}^{\dfrac{1}{x}}}={{0}^{\infty }}
This is of indeterminate form.
So, we will solve the given expression before applying the limits.
Consider the given expression, it can be written as,
lim⁡x→0+ (sin⁡x)1x=lim⁡x→0+ sin⁡xx\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{(\sin x)}^{\dfrac{1}{x}}}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\sqrt[x]{\sin x}
Multiplying and dividing by xx\sqrt[x]{x}, we get
lim⁡x→0+ (sin⁡x)1x=lim⁡x→0+ sin⁡xx×xxxx ⇒lim⁡x→0+ (sin⁡x)1x=lim⁡x→0+ sin⁡xxxx×xx ⇒lim⁡x→0+ (sin⁡x)1x=lim⁡x→0+ sin⁡xxx×xx \begin{aligned} & \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{(\sin x)}^{\dfrac{1}{x}}}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\sqrt[x]{\sin x}\times \dfrac{\sqrt[x]{x}}{\sqrt[x]{x}} \\\ & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{(\sin x)}^{\dfrac{1}{x}}}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{\sqrt[x]{\sin x}}{\sqrt[x]{x}}\times \sqrt[x]{x} \\\ & \Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{(\sin x)}^{\dfrac{1}{x}}}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\sqrt[x]{\dfrac{\sin x}{x}}\times \sqrt[x]{x} \\\ \end{aligned}
Now we know the limit of the product is the product of the limits. So, the limit of product of two functions is equal to the product of individual limits of the functions, therefore
⇒lim⁡x→0+ (sin⁡x)1x=lim⁡x→0+ sin⁡xxx×lim⁡x→0+ xx\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{(\sin x)}^{\dfrac{1}{x}}}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\sqrt[x]{\dfrac{\sin x}{x}}\times \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\sqrt[x]{x}
Now we know, lim⁡x→0 sin⁡xx=1\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1, so the above equation becomes,
⇒lim⁡x→0+ (sin⁡x)1x=1×lim⁡x→0+ xx\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{(\sin x)}^{\dfrac{1}{x}}}=1\times \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\sqrt[x]{x}
Now the limit says xx is approaching zero from right hand side, that is it includes all the positive numbers, so applying the limits to the above equation, we get
⇒lim⁡x→0+ (sin⁡x)1x=0\Rightarrow \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{(\sin x)}^{\dfrac{1}{x}}}=0
So the limit of the given expression is zero.

Note : Students normally are worried about seeing such a question, despite solving it in minutes. If the limit had x→0−x\to {{0}^{-}} , then we would get a different answer as there is no real root for negative numbers. We will get complex numbers for the roots of negative numbers.