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

The limit of the quotient of $\sqrt{2+\cos{x}}-1$ by $(\pi-x)^2$ has to evaluate as $x$ approaches $\pi$ in this limit problem.

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

Now, evaluate the limit of the quotient of the function $\sqrt{2+\cos{x}}-1$ by another function $(\pi-x)^2$ as $x$ approaches $\pi$ as per direct substitution method.

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

$= \,\,\,$ $\dfrac{\sqrt{2-1}-1}{0^2}$

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

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

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

The direct substitution method is failed to find the limit of the given algebraic trigonometric function as $x$ approaches $\pi$. So, we have to think about to find the limit in another mathematical approach.

Convert the function in another variable

1. If $x \to \pi$, then $x-\pi \to 0$.
2. If $x-\pi \to 0$, then $-(\pi-x) \to 0$.
3. If $-(\pi-x) \to 0$, then $(\pi-x) \to 0$.

Therefore, it is derived that if $x$ approaches $\pi$, then $\pi-x$ approaches zero. Take $y = \pi-x$, then $x = \pi-y$. Now, eliminate the $x$ by expressing the whole function in terms of $y$.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sqrt{2+\cos{(\pi-y)}}-1}{y^2}}$

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

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

Simplify the function by the rationalization

If we try the direct substitution method, once again the limit of the function is equal to indeterminate. So, the function has to simplify in another approach. In numerator, the expression is in radical form and it gives us a clue to use rationalisation method.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg(\dfrac{\sqrt{2-\cos{y}}-1}{y^2} \times 1 \Bigg)}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg(\dfrac{\sqrt{2-\cos{y}}-1}{y^2} \times \dfrac{\sqrt{2-\cos{y}}+1}{\sqrt{2-\cos{y}}+1} \Bigg)}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg(\dfrac{(\sqrt{2-\cos{y}}-1) \times (\sqrt{2-\cos{y}}+1)}{y^2 \times (\sqrt{2-\cos{y}}+1)} \Bigg)}$

In numerator of the function, the expression in product form represents the factor form of the difference of the squares formula.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{(\sqrt{2-\cos{y}})^2-(1)^2}{y^2(\sqrt{2-\cos{y}}+1)}}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{2-\cos{y}-1}{y^2(\sqrt{2-\cos{y}}+1)}}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{2-1-\cos{y}}{y^2(\sqrt{2-\cos{y}}+1)}}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{1-\cos{y}}{y^2(\sqrt{2-\cos{y}}+1)}}$

Split the function as product of two functions

Now, split the function as the product of two separate functions.

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg(\dfrac{1-\cos{y}}{y^2} \times \dfrac{1}{\sqrt{2-\cos{y}}+1}\Bigg)}$

Now, use product rule of limits.

$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{1-\cos{y}}{y^2} \Bigg)}$ $\times$ $\Bigg(\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{1}{\sqrt{2-\cos{y}}+1}\Bigg)}$

Evaluate the limit of the function by direct substitution

Don’t disturb the limit of the first function but evaluate the second function by using direct substitution method as $y$ approaches $0$.

$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{1-\cos{y}}{y^2} \Bigg)}$ $\times$ $\Bigg(\dfrac{1}{\sqrt{2-\cos{(0)}}+1}\Bigg)$

$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{1-\cos{y}}{y^2} \Bigg)}$ $\times$ $\Bigg(\dfrac{1}{\sqrt{2-1}+1}\Bigg)$

$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{1-\cos{y}}{y^2} \Bigg)}$ $\times$ $\Bigg(\dfrac{1}{\sqrt{1}+1}\Bigg)$

$= \,\,\,$ $\Bigg(\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{1-\cos{y}}{y^2} \Bigg)}$ $\times$ $\Bigg(\dfrac{1}{1+1}\Bigg)$

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

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

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

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

Now, use the constant multiple rule of limits.

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

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

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

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

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

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

Evaluate the limit of the function

There is a trigonometric function in numerator and also an algebraic function in denominator. The whole function is almost in the form of the limit rule of trigonometric function. So, let’s try to express the function same as the limit rule.

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

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

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{y}{2}\Big)}}{4 \times \dfrac{1}{4} \times y^2}}$

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{y}{2}\Big)}}{4 \times \dfrac{1 \times y^2}{4}}}$

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

$= \,\,\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg(\dfrac{1}{4} \times \dfrac{\sin^2{\Big(\dfrac{y}{2}\Big)}}{\dfrac{y^2}{4}}\Bigg)}$

Once again, use the constant multiple rule of limits.

$= \,\,\,$ $\dfrac{1}{4} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{y}{2}\Big)}}{\dfrac{y^2}{4}}}$

$= \,\,\,$ $\dfrac{1}{4} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{y}{2}\Big)}}{\dfrac{y^2}{2^2}}}$

$= \,\,\,$ $\dfrac{1}{4} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{y}{2}\Big)}}{\Big(\dfrac{y}{2}\Big)^2}}$

Use, the quotient rule of exponents with same base.

$= \,\,\,$ $\dfrac{1}{4} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin{\Big(\dfrac{y}{2}\Big)}}{\Big(\dfrac{y}{2}\Big)}\Bigg)^2}$

Now, use the limit rule of an exponential function.

$= \,\,\,$ $\dfrac{1}{4} \times \Bigg(\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{y}{2}\Big)}}{\Big(\dfrac{y}{2}\Big)}\Bigg)^2}$

If $y$ approaches $0$, then $\dfrac{y}{2} \to \dfrac{0}{2}$. Therefore, $\dfrac{y}{2}$ also approaches $0$.

$= \,\,\,$ $\dfrac{1}{4} \times \Bigg(\displaystyle \large \lim_{\Large \frac{y}{2} \large \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{y}{2}\Big)}}{\Big(\dfrac{y}{2}\Big)}\Bigg)^2}$

Take $p = \dfrac{y}{2}$ and express the whole function in terms of $p$.

$= \,\,\,$ $\dfrac{1}{4} \times \Bigg(\displaystyle \large \lim_{p \,\to\, 0}{\normalsize \dfrac{\sin{p}}{p}\Bigg)^2}$

According to the limit sinx/x as x approaches 0 rule, the limit of the quotient of $\sin{p}$ by $p$ as $x$ approaches $0$ is also equal to $1$.

$= \,\,\,$ $\dfrac{1}{4} \times \Big(1\Big)^2$

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

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

$= \,\,\,$ $0.25$

Therefore, it is evaluated in calculus that the limit of the quotient of $\sqrt{2+\cos{x}}-1$ by $(\pi-x)^2$ as $x$ approaches $\pi$ is equal to $\dfrac{1}{4}$ or $0.25$.

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