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