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Model Paper 5 — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSModel Paper 53 Marks
Question
Evaluate $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)$
✓

Answer

To evaluate: $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)$
Formula used: 
L'Hospital's rule 
Let f(x) and g(x) be two functions which are differentiable on an open interval I except at a point a where 
$\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} g(x)=0$ or $\pm \infty$ then $\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$
As $x \rightarrow 0$, we have
$\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\frac{0}{0}$
This represents an indeterminate form. Thus applying L'Hospital's rule, we get 
$\begin{array}{l}\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\lim _{x \rightarrow 2} \frac{\frac{d}{d x}\left(x^2-4\right)}{\frac{d}{d x}(\sqrt{x+2}-\sqrt{3 x-2})} \\ \lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\lim _{x \rightarrow 2} \frac{2 x}{\frac{1}{2 \sqrt{x+2}}-\frac{3}{2 \sqrt{x-2}}} \\ \lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\frac{4}{\frac{1}{2 \sqrt{2+2}}-\frac{3}{2 \sqrt{5-2}}} \\ \lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=\frac{8}{\frac{1}{2}-\frac{3}{2}} \\ \lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)=-8\end{array}$
Thus, the value of $\lim _{x \rightarrow 2}\left(\frac{x^2-4}{\sqrt{x+2}-\sqrt{3 x-2}}\right)$ is -8

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