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Consider the following expression.$$\displaystyle \lim_{x\rightarrow-3}\frac{\sqrt{2x+22}-4}{x+3}$$The value of the above expression (rounded to 2 decimal places) is ___________.
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We can apply L’Hopital’s rule. Ans is 0.25

Copy paste from Ruturaj sir’s mock test.

https://gateoverflow.in/285459/go2019-flt1-19

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For square root of 16, isn’t -4 also possible?
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Best answer
$\displaystyle \lim_{x \to -3} \frac{\sqrt{2x+22}\,-4}{x+3}\; (\frac{0}{0}\,form)$

$\text{Using L'Hôpital's rule}$

$\displaystyle \lim_{x \to -3} \frac{\frac{1}{2\sqrt{2x+22}}(2)\,-0}{1+0} =\lim_{x \to -3}\frac{1}{\sqrt{2x+22}} =\frac{1}{\sqrt{2(-3)+22}} =\frac{1}{4}=0.25$
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ANS IS 0.25

BY APPLYING L’HOSPITAL RULE WE CAN EASILY GET THIS https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

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from where you study calculus for gate please tell resources
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