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# Evaluate $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi {{\cos }^{2}}x \right)}{{{x}^{2}}}$ is equal to. A. $-\pi$ B. $\pi$ C. $\dfrac{\pi }{2}$D. 1

Last updated date: 21st Mar 2023
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Hint: Put ${{\cos }^{2}}x=1-{{\sin }^{2}}x.$ Then modify the function as limit of trigonometric function. Multiply by $\left( \pi {{\sin }^{2}}x \right)$in numerator and denominator.

In this limit we have a trigonometric question where the variables x represent the angle of the right angle triangle.
First simplify the trigonometric function by identities,
$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi {{\cos }^{2}}x \right)}{{{x}^{2}}}...................\left( i \right)$
We know ${{\cos }^{2}}x+{{\sin }^{2}}x=1$
$\Rightarrow {{\cos }^{2}}x=1-{{\sin }^{2}}x.$
Substitute value of ${{\cos }^{2}}x$ in equation (i);
$=\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( n\left( 1-{{\sin }^{2}}x \right) \right)}{{{x}^{2}}}-\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi -\pi {{\sin }^{2}}x \right)}{{{x}^{2}}}$
We know it’s of the form $\sin \left( A-B \right)=\sin A\cos B-\cos A\sin B$
$\therefore \sin \left( \pi -\pi {{\sin }^{2}}x \right)=\left( \sin \pi \times \cos \left( \pi {{\sin }^{2}}x \right) \right)-\left( \cos \pi \sin \left( \pi {{\sin }^{2}}x \right) \right)$
We know $\sin \pi =0\ \And \,\cos \pi =-1$
$\therefore \sin \left( \pi -\pi {{\sin }^{2}}x \right)=\sin \left( \pi {{\sin }^{2}}x \right)$
The angle including the sine function belongs to the 2nd quadrant. The sine function is positive in the 2nd quadrant.
\begin{align} & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin \pi {{\sin }^{2}}x}{{{x}^{2}}} \\ & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{{{x}^{2}}} \\ \end{align}
$\Rightarrow \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{{{x}^{2}}}$, sine function is positive in second quadrant.
Now we have to modify the function as the limit of the trigonometric function. Remember that if a sine function involves a limit, then you must try to transform the function exactly as the limit of the quotient of sin x by x, as x approaches zero rule.
$\therefore$ Multiply and divide the function $\pi {{\sin }^{2}}x$.
\begin{align} & =\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{{{x}^{2}}}\times 1 \right] \\ & =\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{{{x}^{2}}}\times \dfrac{\pi {{\sin }^{2}}x}{\pi {{\sin }^{2}}x} \right] \\ & =\underset{x\to 0}{\mathop{\lim }}\,\left[ \dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{{{x}^{2}}}\times \dfrac{\pi {{\sin }^{2}}x}{{{x}^{2}}} \right] \\ \end{align}
Apply the product rule of limits, the limit of product of two functions is equal to product of their limits.
\begin{align} & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{\pi {{\sin }^{2}}x}\times \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\pi {{\sin }^{2}}x}{{{x}^{2}}} \\ & =\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{\pi {{\sin }^{2}}x}\times \pi \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{2}}x}{{{x}^{2}}} \\ \end{align}
$\pi {{\sin }^{2}}x$is the angle inside the sine function and its denominator but the same angle should be the input for the first limit function.
We know $x\to 0$,
$\sin x\to \sin \left( 0 \right)\Rightarrow \sin x\to 0$
Similarly ${{\sin }^{2}}x\to {{0}^{2}}\ \ \therefore {{\sin }^{2}}\to 0$
$\pi {{\sin }^{2}}x\to \pi \times 0\ \ \ \therefore \pi {{\sin }^{2}}x\to 0$
$\therefore$ if $x\to 0$, then $\pi {{\sin }^{2}}x\to 0$
\begin{align} & \Rightarrow \underset{\pi {{\sin }^{2}}x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{\pi {{\sin }^{2}}x}\times \pi \underset{x\to 0}{\mathop{\lim }}\,\dfrac{{{\sin }^{2}}x}{{{x}^{2}}} \\ & =\underset{\pi {{\sin }^{2}}x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{\pi {{\sin }^{2}}x}\times \pi \underset{x\to 0}{\mathop{\lim }}\,{{\left( \dfrac{\sin x}{x} \right)}^{2}} \\ & =\underset{\pi {{\sin }^{2}}x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{\pi {{\sin }^{2}}x}\times \pi {{\left( \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x} \right)}^{2}} \\ \end{align}
Use the limit of $\dfrac{\sin x}{x}$ rule as x approaches 0.
The limit of $\dfrac{\sin x}{x}$ as $x\to 0$is equal to 1 and apply it to each function to solve this limit trigonometric problem.
\begin{align} & \therefore \underset{\pi {{\sin }^{2}}x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{\pi {{\sin }^{2}}x}=1 \\ & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}=1 \\ & \Rightarrow \underset{\pi {{\sin }^{2}}x\to 0}{\mathop{\lim }}\,\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{\pi {{\sin }^{2}}x}\times \pi {{\left[ \underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x} \right]}^{2}} \\ & 1\times \pi {{\left( 1 \right)}^{2}}=1\times \pi \\ & =\pi \\ \end{align}
Therefore, it successfully solved the limit and the answer is option B.

Note: Each function in limit form is almost similar to $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\sin x}{x}$, but it is essential to make adjustments for applying limit trigonometric rule.