Question

solve it (Ans-1/8)..,.....

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Answer 1

$\huge\mathbb{\red{GIVEN:}}$

★$\displaystyle\lim_{x\to1} \frac{4-\sqrt{15+x}}{1-x}$

$\huge\mathbb{\red {TO\:\:FIND:}}$

★The value of the given limit.

$\huge\mathbb{\red {CONCEPT\:\:USED:}}$

★We will use 'Method of Rationalisation ' for simplifying the limit.

$\huge\mathbb{\red {ANSWER:}}$

We have,

=$\displaystyle\lim_{x\to1} \frac{4-\sqrt{15+x}}{1-x}$

=$\displaystyle\lim_{x\to1}\frac{(4-\sqrt{15+x}) (4+\sqrt{15+x})}{(1-x) (4+\sqrt{15+x})}$

=$\displaystyle\lim_{x\to1}\frac{ (4) ^{2}-(\sqrt{15+x}) ^{2}}{(1-x) (4+\sqrt{15+x}) }$

$\large\purple{\boxed{\orange{(a+b) (a-b) =a^2-b^2}}}$

=$\displaystyle\lim_{x\to1}\frac{16-15-x}{(1-x) (4+\sqrt{15+x}) }$

=$\displaystyle\lim_{x\to1}\frac{(1-x)}{(1-x) (4+\sqrt{15+x}) }$

=$\displaystyle\lim_{x\to1}\frac{\cancel{(1-x)}}{\cancel{(1-x)} (4+\sqrt{15+x}) }$

=$\displaystyle\lim_{x\to1}\frac{1}{4+\sqrt{15+x}}$

Now putting x = 1 , we have

=$\dfrac{1}{4+\sqrt{15+1}}$

=$\dfrac{1}{4+\sqrt{16}}$

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

=$\frac{1}{8}$

The function approaches the value $\frac{1}{8}$.

$<marquee scrollamount="1300">♥Answer by Rishabh♥</marquee>$