Question

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$\small{\bold \red{ \displaystyle \lim_{x \to - 2}( \dfrac{ \sqrt{x + 6} - 2}{x + 2})}}$

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Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate the given limit.

$=\displaystyle\sf\lim_{x\to-2}\bigg(\dfrac{\sqrt{x+6}-2}{x+2}\bigg)$

On evaluating the given limit, we will get an indeterminate form. So, we have to use L'Hospital's rule.

Applying L'Hospital's rule, we get:

$=\displaystyle\sf\lim_{x\to-2}\bigg(\dfrac{\dfrac{d}{dx}(\sqrt{x+6}-2)}{\dfrac{d}{dx}(x+2)}\bigg)$

$=\displaystyle\sf\lim_{x\to-2}\bigg(\dfrac{\dfrac{d}{dx}(\sqrt{x+6})-\dfrac{d}{dx}(2)}{\dfrac{d}{dx}(x)+\dfrac{d}{dx}(2)}\bigg)$

As we know that:

$\sf:\longmapsto \dfrac{d}{dx}(\sqrt{x})=\dfrac{1}{2\sqrt{x}}$

$\sf:\longmapsto \dfrac{d}{dx}(k)=0$

$\sf:\longmapsto \dfrac{d}{dx}(x^{n})=nx^{n-1}$

We get:

$=\displaystyle\sf\lim_{x\to-2}\bigg(\dfrac{\dfrac{1}{2\sqrt{x+6}}}{1}\bigg)$

$=\displaystyle\sf\lim_{x\to-2}\bigg(\dfrac{1}{2\sqrt{x+6}}\bigg)$

Now put x = -2 in the given expression, we get:

$=\sf\dfrac{1}{2\sqrt{-2+6}}$

$=\sf\dfrac{1}{2\sqrt{4}}$

$=\sf\dfrac{1}{2\cdot 2}$

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

Therefore:

$\displaystyle\sf:\longmapsto\lim_{x\to-2}\bigg(\dfrac{\sqrt{x+6}-2}{x+2}\bigg)=\dfrac{1}{4}$

$\textsf{\large{\underline{Additional Information}:}}$

$\begin{gathered}\boxed{\begin{array}{c|c}\bf f(x)&\bf\dfrac{d}{dx}f(x)\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf0\\ \\ \sf sin(x)&\sf cos(x)\\ \\ \sf cos(x)&\sf-sin(x)\\ \\ \sf tan(x)&\sf{sec}^{2}(x)\\ \\ \sf cot(x)&\sf-{cosec}^{2}(x)\\ \\ \sf sec(x)&\sf sec(x)tan(x)\\ \\ \sf cosec(x)&\sf-cosec(x)cot(x)\\ \\ \sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\\ \\ \sf log(x)&\sf\dfrac{1}{x}\\ \\ \sf{e}^{x}&\sf{e}^{x}\end{array}}\\ \end{gathered}$