# 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

Answer

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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.

Complete step-by-step answer:

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 itâ€™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.

Complete step-by-step answer:

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 itâ€™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.

Last updated date: 02nd Oct 2023

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