# 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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