Evaluate $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{x}\Big)}}{x-2}}$

A trigonometric function $\cos{\Big(\dfrac{\pi}{x}\Big)}$ and an algebraic function $x-2$ formed a fractional function by division. The limit of this algebraic trigonometric function has to evaluate as $x$ approaches $2$ in this trigonometric limit problem.

Let us try to evaluate the limit of the trigonometric function by the direct substitution method as $x$ approaches $2$.

$= \,\,\,$ $\dfrac{\cos{\Big(\dfrac{\pi}{2}\Big)}}{2-2}$

The cosine of angle $90$ degrees is zero as per trigonometry.

$= \,\,\,$ $\dfrac{0}{0}$

It is an indeterminate form. So, the direct substitution method is failed in calculating the limit of the given function as $x$ approaches $2$. So, we have to evaluate the limit of the function in another approach. The trigonometric limit problem can be done in two methods.

Method – 1

It is given that $x \,\to\, 2$, then $x-2 \,\to\, 2-2$. Therefore, $x-2 \,\to\, 0$. It clears that $x-2$ approaches zero when $x$ approaches $2$.

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

Convert the whole function in another variable

Now, take $y = x-2$, then $x = y+2$ mathematically. Finally, transform the entire function in terms of $y$ from $x$.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{y+2}\Big)}}{y}}$

You will get indeterminate form if you try to evaluate this trigonometric function as $y$ approaches zero by the direct substitution method. So, we have to take a different step here.

Express the cos function in terms of sine

In calculus, there is no trigonometric limit rule in cosine but we have a trigonometric limit rule in sine. So, we have to express the cosine function in terms of sine for moving ahead in simplification. It can be done by the cofunction identity of sine.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{y+2}\Big)}}{y}}$

Simplify the Algebra trigonometric function

The algebraic trigonometric function is almost same as the trigonometric limit rule of sine function. So, let us try to adjust the whole function for transforming the current function into required form.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi(y+2)-\pi(2)}{2(y+2)}\Big)}}{y}}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y+2\pi-2\pi}{2(y+2)}\Big)}}{y}}$

$= \,\,\,$ $\require{cancel} \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y+\cancel{2\pi}-\cancel{2\pi}}{2(y+2)}\Big)}}{y}}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{y}}$

The denominator of the fractional function should be same as the angle in the sine function for applying the trigonometric limit rule of sine function in this problem. Hence, let us try to adjust the denominator of the function by taking some acceptable mathematical steps.

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

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

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

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

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

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

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

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

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

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

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

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

Evaluate the Limit of the function

A fractional function is split as the product of two fractional functions. The limit of the product of them can be evaluated by calculating the product of their limits. So, use the product rule of limits.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\pi}{2(y+2)}}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$

The limit of first fractional function can be evaluated by the direct substitution method but do not disturb the limit of second fractional function.

$= \,\,\,$ $\dfrac{\pi}{2(0+2)}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$

$= \,\,\,$ $\dfrac{\pi}{2(2)}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$

$= \,\,\,$ $\dfrac{\pi}{2 \times 2}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$

$= \,\,\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$

The function in the denominator is same as the angle in the sine function. So, the trigonometric limit rule of sine function can be applied.

Now, take $z \,=\, \dfrac{\pi y}{2(y+2)}$. We have to express the input in terms of the $z$ completely in the function.

$(1) \,\,\,$ If $y \,\to\, 0$, then $\pi \times y \,\to\, \pi \times 0$. Therefore, $\pi y \,\to\, 0$.

$(2) \,\,\,$ If $\pi y \,\to\, 0$, then $\dfrac{\pi y}{2} \,\to\, \dfrac{0}{2}$. Therefore, $\dfrac{\pi y}{2} \,\to\, 0$.

$(3) \,\,\,$ If $y \,\to\, 0$, then $y+2 \,\to\, 0+2$. Therefore, $y+2 \,\to\, 2$. If $\dfrac{\pi y}{2} \times \dfrac{1}{y+2} \,\to\, 0 \times \dfrac{1}{y+2}$, then $\dfrac{\pi y}{2(y+2)} \,\to\, 0 \times \dfrac{1}{2}$. Therefore, $\dfrac{\pi y}{2(y+2)} \,\to\, 0$ but we have assumed that $z \,=\, \dfrac{\pi y}{2(y+2)}$.

Therefore, the three steps have cleared that if $y$ approaches zero, then the $z$ also approaches zero. Now, convert the function in terms of $z$ from $y$.

$\implies$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi y}{2(y+2)}\Big)}}{\dfrac{\pi y}{2(y+2)}}}$ $\,=\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{z \,\to\, 0}{\normalsize \dfrac{\sin{z}}{z}}$

According to the trigonometric limit rule of sine function, the limit rule of sinx/x as x approaches 0 is equal to one. Therefore, the limit of $\dfrac{\sin{z}}{z}$ as $z$ approaches $0$ is also equal to one.

$=\,\,\,$ $\dfrac{\pi}{4} \times 1$

$=\,\,\,$ $\dfrac{\pi}{4}$

Method – 2

In this method, we do not use the transformation technique but we take some acceptable mathematical steps, which help us to evaluate the limit of the given function.

$\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\cos{\Big(\dfrac{\pi}{x}\Big)}}{x-2}}$

Express the cos function in terms of sine

In the given function, the numerator is a trigonometric function and the denominator is algebraic function. In limits, there is no limit rule in cosine function but we have a trigonometric limit rule in sine function. Hence, we have to convert the cosine function into sine function and it is possible by the complementary angle identity of sine function.

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

Adjust denominator same as Angle of sine

In order to use the trigonometric limit rule of sine function, we have to adjust the algebraic function in the denominator same as the angle in the sine function.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Evaluate the Limits of the functions

Now, split the limit of the product of functions as product of their limits by the product rule of limits.

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

Now, find the limit of the first fractional function by the direct substitution method as $x$ approaches $2$.

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

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

$=\,\,\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$

Now, let us evaluate the limit of the second trigonometric function. The algebraic function in the denominator is exactly same as the angle in the sine function. So, we can use the trigonometric limit rule of sine function but we have to set the input of the limit same as the denominator or angle in the sine function.

$(1) \,\,\,$ If $x \,\to\, 2$ then $\dfrac{1}{x} \,\to\, \dfrac{1}{2}$

$(2) \,\,\,$ If $\dfrac{1}{x} \,\to\, \dfrac{1}{2}$, then $\pi \times \dfrac{1}{x} \,\to\, \pi \times \dfrac{1}{2}$. Therefore, $\dfrac{\pi}{x} \,\to\, \dfrac{\pi}{2}$

$(3) \,\,\,$ If $\dfrac{\pi}{x} \,\to\, \dfrac{\pi}{2}$, then $-\dfrac{\pi}{x} \,\to\, -\dfrac{\pi}{2}$

$(4) \,\,\,$ If $-\dfrac{\pi}{x} \,\to\, -\dfrac{\pi}{2}$, then $\dfrac{\pi}{2}-\dfrac{\pi}{x} \,\to\, \dfrac{\pi}{2}-\dfrac{\pi}{2}$. So, $\dfrac{\pi}{2}-\dfrac{\pi}{x} \,\to\, 0$

The four steps have cleared to us that $\dfrac{\pi}{2}-\dfrac{\pi}{x}$ approaches zero when $x$ approaches $2$.

Now, take $m = \dfrac{\pi}{2}-\dfrac{\pi}{x}$ and convert the function in terms of $m$ from $x$.

$\implies$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 2}{\normalsize \dfrac{\sin{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}{\Big(\dfrac{\pi}{2}-\dfrac{\pi}{x}\Big)}}$ $\,=\,$ $\dfrac{\pi}{4}$ $\times$ $\displaystyle \large \lim_{m \,\to\, 0}{\normalsize \dfrac{\sin{m}}{m}}$

According to the trigonometric limit rule of sine function, the limit of $\dfrac{\sin{m}}{{m}}$ as $m$ approaches zero is equal to one.

$=\,\,\,$ $\dfrac{\pi}{4} \times 1$

$=\,\,\,$ $\dfrac{\pi}{4}$

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