In this limit trigonometric question, the variable $x$ represents angle of right triangle. The quotient of sin of angle pi times cosine squared $x$ by $x$ squared formed a special function and we need to find the limit of this function as $x$ approaches zero.

$\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{(\pi\cos^2{x})}}{x^2}}$

It’s very important to simplify the trigonometric function before finding its limit as $x$ tends to $0$. It can be done by using two basic trigonometric identities. According to cos squared formula, the square of cos function can be expanded in terms of square of sin function.

$= \,\,\,$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{(\pi{(1-\sin^2{x})})}}{x^2}}$

$= \,\,\,$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{(\pi-\pi\sin^2{x})}}{x^2}}$

Look at the angle inside the sine function and it belongs to second quadrant. The sine function is positive in the second quadrant.

$= \,\,\,$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{(\pi\sin^2{x})}}{x^2}}$

Remember that if sine function involves in limits, then you must try to transform the function exactly same as the limit of quotient of $\sin{x}$ by $x$ as $x$ approaches zero rule. So, multiply and divide the function by the angle inside the sin function. In this case $\pi \sin^2{x}$ is angle inside the sine function.

$= \,\,\,$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\sin{(\pi\sin^2{x})}}{x^2} \times 1 \Bigg]}$

$= \,\,\,$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\sin{(\pi\sin^2{x})}}{x^2} \times \dfrac{\pi\sin^2{x}}{\pi\sin^2{x}} \Bigg]}$

$= \,\,\,$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \Bigg[ \dfrac{\sin{(\pi\sin^2{x})}}{\pi\sin^2{x}} \times \dfrac{\pi\sin^2{x}}{x^2} \Bigg]}$

As per Product Rule of Limits, the limit of product of two functions is equal to product of their limits.

$= \,\,\,$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{(\pi\sin^2{x})}}{\pi\sin^2{x}}}$ $\times$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\pi\sin^2{x}}{x^2}}$

In the second factor, a constant (pi) is multiplying a function. It can be simplified by using constant multiple rule

$= \,\,\,$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{(\pi\sin^2{x})}}{\pi\sin^2{x}}}$ $\times$ $\pi \large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$

$= \,\,\,$ $\pi \large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{(\pi\sin^2{x})}}{\pi\sin^2{x}}}$ $\times$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2}}$

Each function in limit form is almost similar to the limit of ratio of $\sin{x}$ to $x$ as $x$ tends to $0$ rule but it is essential to make some adjustments for applying limit trigonometric rule.

$\pi \sin^2{x}$ is angle inside the sine function and it’s also in denominator but the same angle should be the input for the first limit function. In other words, if $x \to 0$, then $\sin{x} \to \sin{(0)}$. Hence, $\sin{x} \to 0$. Similarly, $\sin^2{x} \to {(0)}^2$. So, $\sin^2{x} \to 0$. Now, $\pi \times \sin^2{x} \to \pi \times 0$. Therefore, $\pi \sin^2{x} \to 0$

Therefore, if $x$ approaches $0$, then $\pi \sin^2{x}$ also approaches to $0$. You can now change the input in the first limit function but no need to change in second function.

$= \,\,\,$ $\pi$ $\Bigg[ \large \displaystyle \lim_{\pi \sin^2{x} \,\to\, 0}{\normalsize \dfrac{\sin{(\pi\sin^2{x})}}{\pi\sin^2{x}}}$ $\times$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin^2{x}}{x^2} \Bigg]}$

It’s time to transform the second multiplying factor same as the sine rule in limits.

$= \,\,\,$ $\pi$ $\Bigg[ \large \displaystyle \lim_{\pi \sin^2{x} \,\to\, 0}{\normalsize \dfrac{\sin{(\pi\sin^2{x})}}{\pi\sin^2{x}}}$ $\times$ $\large \displaystyle \lim_{x \,\to\, 0}{\normalsize {\Big( \dfrac{\sin{x}}{x} \Big)}^2 \Bigg]}$

The second multiplying function can be further simplified by using power rule of limits.

$= \,\,\,$ $\pi$ $\Bigg[ \large \displaystyle \lim_{\pi \sin^2{x} \,\to\, 0}{\normalsize \dfrac{\sin{(\pi\sin^2{x})}}{\pi\sin^2{x}}}$ $\times$ ${\Big( \large \displaystyle \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{x}}{x} \Big)}^2}\Bigg]$

The limit of sinx by x as x approaches 0 is equal to one and apply it to each function to solve this limit trigonometric problem.

$= \,\,\,$ $\pi$ $[1 \times {(1)}^2]$

$= \,\,\,$ $\pi$ $[1 \times 1]$

$= \,\,\,$ $\pi \times 1$

$= \,\,\,$ $\pi$

Therefore, it’s successfully solved that the limit of quotient of sine of pi times cos squared $x$ by $x$ squared is equal to $\pi$ as $x$ approaches zero.

Latest Math Topics

Latest Math Problems

A best free mathematics education website for students, teachers and researchers.

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Learn how to solve the maths problems in different methods with understandable steps.

Copyright © 2012 - 2021 Math Doubts, All Rights Reserved