Question

The value of $\displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi {{\cos }^{2}}x \right)}{{{x}^{2}}}$ is equal to:
(a) $\dfrac{\pi }{2}$
(b) $1$
(c) $-\pi$
(d) $\pi$

Answer 1

First of all, we are going to see whether the limit expression is the form of $\dfrac{0}{0}$ or $\dfrac{\infty }{\infty }$. This can be achieved by putting x as 0 in the given expression and see what is coming out from it. Then, we are going to use the trigonometric identity which states that: ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ . Also, we are going to use the trigonometric identity which is equal to: $\sin \left( \pi -\theta \right)=\sin \theta$. Then we are going to use the limit property which says that: $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$ so we are going to rearrange the given expression in such a manner so that this limit property is satisfying.

Complete step by step answer:
The expression of the limit given in the above problem is as follows:
$\displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi {{\cos }^{2}}x \right)}{{{x}^{2}}}$
Substituting the limit value (x as 0) in the above expression and we get,
$\begin{aligned} & \displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi {{\cos }^{2}}\left( 0 \right) \right)}{{{\left( 0 \right)}^{2}}} \\\ & =\dfrac{\sin \left( \pi \right)}{0} \\\ \end{aligned}$
$=\dfrac{0}{0}$
As you can see that we are getting $\dfrac{0}{0}$ form after putting the limit values in the limit expression.
In the above expression, we can write ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ because this is the trigonometric identity. Using this relation, in the above we get,

& \displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi \left( 1-{{\sin }^{2}}x \right) \right)}{{{x}^{2}}} \\\ & =\displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi -\pi {{\sin }^{2}}x \right)}{{{x}^{2}}} \\\ \end{aligned}$$ Now, we know the trigonometric angle conversion in sine as follows: $\sin \left( \pi -\theta \right)=\sin \theta $ Substituting $\theta =\pi {{\sin }^{2}}x$ in the above equation we get, $\sin \left( \pi -\pi {{\sin }^{2}}x \right)=\sin \left( \pi {{\sin }^{2}}x \right)$ Using the above relation in the above limit we get, $$\begin{aligned} & \Rightarrow \displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi -\pi {{\sin }^{2}}x \right)}{{{x}^{2}}} \\\ & =\displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{{{x}^{2}}} \\\ \end{aligned}$$ Now, we are going to rearrange the above expression in such a manner so that the limit is in the following form: $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}$ Multiplying $\pi {{\sin }^{2}}x$ in the numerator and the denominator of the above expression and we get, $\begin{aligned} & \displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)\pi {{\sin }^{2}}x}{{{x}^{2}}\left( \pi {{\sin }^{2}}x \right)} \\\ & =\displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{\left( \pi {{\sin }^{2}}x \right)}\times \dfrac{\pi {{\sin }^{2}}x}{{{x}^{2}}} \\\ \end{aligned}$ In the above expression, $\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{\pi {{\sin }^{2}}x}=1$ and $\dfrac{{{\sin }^{2}}x}{{{x}^{2}}}=1$ so using these relations in the above expression we get, $=\pi $ From the above solution, we have evaluated the limit as $\pi $. **So, the correct answer is “Option d”.** **Note:** To solve the above problem, you must know how to solve the limit expressions which are in the $\dfrac{0}{0}$ form. Then you should know if the $\dfrac{\sin x}{x}$ form is given and the value of x is 0 in it then the value of the expression is 1. Also, you should know some trigonometric identities and properties.