Question

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

Answer 1

Hint: First, before proceeding for this, we must know the some of the basic formulas of trigonometry so that we can change the given terms into other terms as $ {{\cos }^{2}}x=1-{{\sin }^{2}}x $ . Then, again by using the trigonometry rules which says that sine of any angle is positive in second quadrant where $ \theta $ is any angle less than $ {{90}^{\circ }} $ as $ \sin \left( \pi -\theta \right)=\sin \theta $ . Then, dividing and multiplying the whole expression by $ \pi {{\sin }^{2}}x $ to get the solvable form and applying the property of limits for sine function, we get the desired answer.

Complete step-by-step answer:
In this question, we are supposed to find the value of the limit given as $ \displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi {{\cos }^{2}}x \right)}{{{x}^{2}}} $ .
So, before proceeding for this, we must know the some of the basic formulas of trigonometry so that we can change the given terms into other terms as:
 $ {{\cos }^{2}}x=1-{{\sin }^{2}}x $
Then, by substituting this value in the above limit expression, we get:
 $ \displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi \left( 1-{{\sin }^{2}}x \right) \right)}{{{x}^{2}}} $
So, by solving the above bracket, we get:
 $ \displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi -\pi {{\sin }^{2}}x \right)}{{{x}^{2}}} $
Now, again by using the trigonometry rules which says that sine of any angle is positive in second quadrant where $ \theta $ is any angle less than $ {{90}^{\circ }} $ so:
 $ \sin \left( \pi -\theta \right)=\sin \theta $
So, by applying the above condition to the limit expression, we get:
 $ \displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{{{x}^{2}}} $
Then, by dividing and multiplying the whole expression by $ \pi {{\sin }^{2}}x $ to get the solvable form as:
 $ \displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{{{x}^{2}}}\times \dfrac{\pi {{\sin }^{2}}x}{\pi {{\sin }^{2}}x} $
Then, by applying the formula for the limit given as:
 $ \displaystyle \lim_{x \to 0}\dfrac{\sin \left( \pi {{\sin }^{2}}x \right)}{\pi {{\sin }^{2}}x}=1 $
Then, we are left with the terms as:
 $ \displaystyle \lim_{x \to 0}\dfrac{\pi {{\sin }^{2}}x}{{{x}^{2}}} $
Now, again by applying the same property gives us the value as:
 $ \displaystyle \lim_{x \to 0}\dfrac{{{\sin }^{2}}x}{{{x}^{2}}}=1 $
So, we are only left with the value of the function as $ \pi $ .

So, the correct answer is “Option (d)”.

Note: Now, to solve these types of the questions we need to know some of the basic formulas of the limit of the sine function to get the answer accurately and appropriately. So, the basic formula for any function f(x) in sine function form gives the result as:
 $ \displaystyle \lim_{x \to 0}\dfrac{\sin \left( f\left( x \right) \right)}{f\left( x \right)}=1 $ .
So, we did the same in the above question to get the function in this form and then solved it.