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PART - 2 CH - 12 Limits and Derivatives — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSPART - 2 CH - 12 Limits and Derivatives2 Marks
Question
Prove that $\lim _{x \rightarrow 1} \frac{\sqrt{4+x}-\sqrt{5}}{x-1}=\frac{\sqrt{5}}{10}.$
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Answer


$\begin{array}{l}\text { L.H.S. }=\lim _{x \rightarrow 1} \frac{\sqrt{4+x}-\sqrt{5}}{x-1} \\ =\lim _{x \rightarrow 1} \frac{(\sqrt{4+x}-\sqrt{5})(\sqrt{4+x}+\sqrt{5})}{(x-1)(\sqrt{4+x}+\sqrt{5})} \\ =\lim _{x \rightarrow 1} \frac{4+x-5}{(x-1)(\sqrt{4+x}+\sqrt{5})} \\ =\lim _{x \rightarrow 1} \frac{x-1}{(x-1)(\sqrt{4+x}+\sqrt{5})} \\ =\lim _{x \rightarrow 1} \frac{1}{\sqrt{4+x}+\sqrt{5}} \\ =\frac{1}{\sqrt{4+1}+\sqrt{5}}=\frac{1}{\sqrt{5}+\sqrt{5}}=\frac{1}{2 \sqrt{5}} \\ \Rightarrow=\frac{\sqrt{5}}{2 \sqrt{5} \times \sqrt{5}}=\frac{\sqrt{5}}{10}=\text { R.H.S. }\end{array}$

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