Question

Evaluate the limit:............​

Answer 1

$\sf{\displaystyle\lim _{x\to \sqrt{2}}\left(\frac{x^2-2}{x^2+\sqrt{2}x-4}\right)}$

• Apply L'Hopital's Theorem

$\sf{\displaystyle\mathrm{For}\:\lim _{x\to a}\left(\frac{f\left(x\right)}{g\left(x\right)}\right),\:\mathrm{if}\:\lim _{x\to a}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{0}{0}\quad \mathrm{or}\quad \lim _{x\to \:a}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\frac{\pm \infty }{\pm \infty },\:\mathrm{then}}$

$\bf{\displaystyle\lim _{x\to a}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=\lim _{x\to a}\left(\frac{f\:^'\left(x\right)}{g\:^'\left(x\right)}\right)}$

$\sf{\displaystyle=\lim _{x\to \sqrt{2}}\left(\frac{2x}{2x+\sqrt{2}}\right)}$

$\sf{\displaystyle=\lim _{x\to \sqrt{2}}\left(\frac{\sqrt{2}x}{\sqrt{2}x+1}\right)}$

• Plug in the Value x = √2

$\sf{\displaystyle=\frac{\sqrt{2}\sqrt{2}}{\sqrt{2}\sqrt{2}+1}}$

$\sf{\displaystyle =\frac{2}{\sqrt{2}\sqrt{2}+1}}$

$\sf{=\dfrac{2}{3}}$

Answer 2

Explanation:

$\tt \: lim_{x \rightarrow \sqrt{2} }( \dfrac{x {}^{2} - 2}{x {}^{2} + \sqrt{2x} - 4} ) \: \red \bigstar \\ \\ { \underline{\bold{Applying \: L- hospital \: rule,}}} \\ \\ \tt \longrightarrow \: lim_{x \rightarrow \sqrt{2}}( \dfrac{ \dfrac{d}{dx} (x {}^{2} - 2)}{ \dfrac{d}{dx} (x {}^{2} + \sqrt{2x} - 4) } ) \\ \\ \longrightarrow \tt lim_{x \rightarrow \sqrt{2}}( \frac{2x}{2x + \sqrt{2} } ) \\ \\ \longrightarrow \tt( \frac{2 \times \sqrt{2} }{2 \times \sqrt{2} + \sqrt{2} } ) \\ \\ \longrightarrow \tt( \dfrac{2 \sqrt{2} }{3 \sqrt{2} } ) \\ \\ \longrightarrow \tt( \dfrac{2 \cancel{{\sqrt{2}}} }{3 \cancel{\sqrt{2}} } ) \\ \\ \longrightarrow \tt \dfrac{2}{3} \: \green \bigstar$

Hence,

$\bull \: \boxed{\tt \: lim_{x \rightarrow \sqrt{2} }( \dfrac{x {}^{2} - 2}{x {}^{2} + \sqrt{2x} - 4} ) = \dfrac{2}{3}}$

[ Note :- Refer to the attachment for important limits formulae. ]