# Find the value

Question:

Evaluate

$\lim _{x \rightarrow 4}\left(\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right)$

Solution:

To evaluate:

$\lim _{x \rightarrow 4}\left(\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right)$

Formula used:

Multiplying numerator and denominator with conjugates of numerator and denominator i.e

$(1+\sqrt{5-x})(3+\sqrt{5+x})$

$\lim _{x \rightarrow 4}\left(\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right)=\lim _{x \rightarrow 4}\left(\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right)\left(\frac{1+\sqrt{5-x}}{1+\sqrt{5-x}}\right)\left(\frac{3+\sqrt{5+x}}{3+\sqrt{5+x}}\right)$

$\lim _{x \rightarrow 4}\left(\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right)=\lim _{x \rightarrow 4}\left(\frac{4-x}{x-4}\right)\left(\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}\right)$

$\lim _{x \rightarrow 4}\left(\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right)=\lim _{x \rightarrow 4}-\left(\frac{1+\sqrt{5-x}}{3+\sqrt{5+x}}\right)$

$\lim _{x \rightarrow 4}\left(\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right)=-\frac{1}{3}$

Thus, the value of $\lim _{x \rightarrow 4}\left(\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}\right)$ is $-\frac{1}{3}$

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