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Evaluate $\displaystyle \large \lim_{x \,\to\, \frac{\pi}{2}}{\normalsize \dfrac{1+\cos{2x}}{(\pi-2x)^2}}$

In this trigonometric limit problem, $x$ is a variable but represents an angle of a right triangle. The limit of the algebraic trigonometric function $\dfrac{1+\cos{2x}}{(\pi-2x)^2}$ can be evaluated as $x$ approaches $\dfrac{\pi}{2}$ by the direct substitution method.

$= \,\,\,$ $\dfrac{1+\cos{2\Big(\dfrac{\pi}{2}\Big)}}{\Big(\pi-2\Big(\dfrac{\pi}{2}\Big)\Big)^2}$

$= \,\,\,$ $\dfrac{1+\cos{\Big(\dfrac{2\pi}{2}\Big)}}{\Big(\pi-\Big(\dfrac{2\pi}{2}\Big)\Big)^2}$

$= \,\,\,$ $\require{cancel} \dfrac{1+\cos{\Big(\dfrac{\cancel{2}\pi}{\cancel{2}}\Big)}}{\Big(\pi-\dfrac{\cancel{2}\pi}{\cancel{2}}\Big)^2}$

$= \,\,\,$ $\dfrac{1+\cos{\pi}}{(\pi-\pi)^2}$

$= \,\,\,$ $\dfrac{1+(-1)}{0^2}$

$= \,\,\,$ $\dfrac{1-1}{0}$

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

According to the direct substitution method, the limit of the given function $\dfrac{1+\cos{2x}}{(\pi-2x)^2}$ is determinate as $x$ tends to $\dfrac{\pi}{2}$. The direct substitution method is failed in this case. So, it should be evaluated in another mathematical approach.

Transform the variable into equivalent form

In this step, we eliminate the variable $x$ from the given function by the appropriate replacement.

$\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{1+\cos{2x}}{(\pi-2x)^2}}$.

According to the observation of the given function, $2x$ is an angle in the trigonometric function cosine in the numerator and $2x$ is also a term in the denominator. So, try to replace the $2x$ by its equivalent value.

Take $u = \pi-2x$, then $2x = \pi-u$

Therefore, $2x$ can be replaced by $\pi-u$ in the given function to eliminate the $x$ from the function.

Now, let’s change the input of the limit function.

It is given that $x \,\to\, \dfrac{\pi}{2}$ but $2x = \pi-u$. So, $x = \dfrac{\pi-u}{2}$

Therefore, $\dfrac{\pi-u}{2} \,\to\, \dfrac{\pi}{2}$

$\implies$ $-\Big(\dfrac{u-\pi}{2}\Big) \,\to\, \dfrac{\pi}{2}$

$\implies$ $\dfrac{u-\pi}{2} \,\to\, -\dfrac{\pi}{2}$

$\implies$ $\dfrac{u}{2}-\dfrac{\pi}{2} \,\to\, -\dfrac{\pi}{2}$

$\implies$ $\dfrac{u}{2} \,\to\, -\dfrac{\pi}{2}+\dfrac{\pi}{2}$

$\implies$ $\require{cancel} \dfrac{u}{2} \,\to\, -\cancel{\dfrac{\pi}{2}}+\cancel{\dfrac{\pi}{2}}$

$\implies$ $\dfrac{u}{2} \,\to\, 0$

$\implies$ $u \,\to\, 2 \times 0$

$\implies$ $u \,\to\, 0$

Therefore, it is proved that if $x$ approaches $\dfrac{\pi}{2}$, then $u$ approaches zero.

Simplify the Algebraic trigonometric function

According to the previous step, the limit of the given function can be converted in terms of $u$ from $x$ in two different simplifying methods. You can follow any one of them.

Method: 1

In this method, substitute $\pi-2x = u$ and $2x = \pi-u$

$\implies$ $\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{1+\cos{2x}}{(\pi-2x)^2}}$ $\,=\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{1+\cos{(\pi-u)}}{u^2}}$

The angle $\pi-u$ represents an angle in the second quadrant. So, $\cos{(\pi-u)} \,=\, \cos{u}$.

$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{1-\cos{u}}{u^2}}$

According to power reduction trigonometric identity, $1-\cos{\theta} \,=\, 2\sin^2{\Big(\dfrac{\theta}{2}\Big)}$

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

Method: 2

According to power reducing trigonometric identity, the expression in numerator of the given function can be written in its equivalent form. We know that $1+\cos{2\theta} \,=\, 2\cos^2{\theta}$.

$\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{1+\cos{2x}}{(\pi-2x)^2}}$.

$=\,\,\,$ $\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{2\cos^2{x}}{(\pi-2x)^2}}$.

In this method, substitute $\pi-2x = u$ and $x = \dfrac{\pi-u}{2}$. Similarly, it is already proved that if $x \to \dfrac{\pi}{2}$, then $u \to 0$.

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

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

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

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

The angle $\dfrac{\pi}{2}-\dfrac{u}{2}$ belongs to first quadrant. Therefore, $\cos{\Big(\dfrac{\pi}{2}-\dfrac{u}{2}\Big)} \,=\, \sin{\Big(\dfrac{u}{2}\Big)}$

$=\,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2 {\Bigg(\sin{\Big(\dfrac{u}{2}\Big)}\Bigg)}^2 }{u^2}}$.

$=\,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{u}{2}\Big)}}{u^2}}$.

In both methods, we have simplified that $\displaystyle \large \lim_{x \,\to\, \Large \frac{\pi}{2}}{\normalsize \dfrac{1+\cos{2x}}{(\pi-2x)^2}}$ $\,=\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{u}{2}\Big)}}{u^2}}$ by taking $u = \pi-2x$ in this calculus problem.

Evaluate the Limit of the given function

The simplification of the given function is completed and now, we can start the procedure for finding the limit mathematically.

$\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{2\sin^2{\Big(\dfrac{u}{2}\Big)}}{u^2}}$

In the numerator of the function, $\dfrac{u}{2}$ is an angle in the sin squared function. So, try to adjust the function in the denominator same as the numerator for using the limit rule of trigonometric function. The factor $2$ multiples the sin function in numerator and it divides the term in the denominator. So, shift it to denominator.

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

$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{u}{2}\Big)}}{1 \times \dfrac{u^2}{2}}}$

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

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

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

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

$= \,\,\,$ $\displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{1}{2} \times \dfrac{\sin^2{\Big(\dfrac{u}{2}\Big)}}{{\Big(\dfrac{u}{2}\Big)}^2 }}$

$= \,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{u \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{u}{2}\Big)}}{{\Big(\dfrac{u}{2}\Big)}^2 }}$

$= \,\,\,$ $\dfrac{1}{2} \times \displaystyle \large \lim_{u \,\to\, 0}{\normalsize {\Bigg(\dfrac{\sin{\Big(\dfrac{u}{2}\Big)}}{\Big(\dfrac{u}{2}\Big)}\Bigg)}^2 }$

According to constant exponent limit rule, the above function can be written as follows.

$= \,\,\,$ $\dfrac{1}{2} \times {\Bigg(\displaystyle \large \lim_{\Large \frac{u}{2} \large \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{u}{2}\Big)}}{\Big(\dfrac{u}{2}\Big)}}\Bigg)}^2$

Take $m = \dfrac{u}{2}$

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

The limit of the function represents the limit of sinx/x as x approaches 0 rule. Therefore, the limit of the $\dfrac{\sin{m}}{m}$ function as $m$ approaches $0$ is equal to one.

$= \,\,\,$ $\dfrac{1}{2} \times {\Big(1\Big)}^2$

$= \,\,\,$ $\dfrac{1}{2} \times 1$

$= \,\,\,$ $\dfrac{1}{2}$

Therefore, it is calculated that the limit of the given function $\dfrac{1+\cos{2x}}{(\pi-2x)^2}$ as $x$ approaches $\dfrac{\pi}{2}$ is equal to the $\dfrac{1}{2}$

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