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.
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