GATE CS 2021 Set 1 Q20 — Calculus: Limits, Continuity, Differentiability

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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Answer: 0.25 : 0.25

To find the value of the limit:

L = \lim_{x\rightarrow-3}\frac{\sqrt{2x+22}-4}{x+3}

First, substitute $x = -3$ into the expression to check for indeterminacy:

Since this is a $\frac{0}{0}$ indeterminate form, we can evaluate it using either L'Hôpital's Rule or by rationalizing the numerator.

Differentiating the numerator and the denominator with respect to $x$ :

\frac{d}{dx}(\sqrt{2x+22}-4) = \frac{1}{2\sqrt{2x+22}} \cdot 2 = \frac{1}{\sqrt{2x+22}}

\frac{d}{dx}(x+3) = 1

Applying the limit:

L = \lim_{x\rightarrow-3} \frac{\frac{1}{\sqrt{2x+22}}}{1} = \frac{1}{\sqrt{2(-3)+22}} = \frac{1}{\sqrt{16}} = \frac{1}{4} = 0.25

Multiply the numerator and the denominator by the conjugate of the numerator, $(\sqrt{2x+22}+4)$ :

L = \lim_{x\rightarrow-3}\frac{(\sqrt{2x+22}-4)(\sqrt{2x+22}+4)}{(x+3)(\sqrt{2x+22}+4)}

L = \lim_{x\rightarrow-3}\frac{(2x+22)-16}{(x+3)(\sqrt{2x+22}+4)}

L = \lim_{x\rightarrow-3}\frac{2(x+3)}{(x+3)(\sqrt{2x+22}+4)}

Canceling the common factor $(x+3)$ for $x \neq -3$ :

L = \lim_{x\rightarrow-3}\frac{2}{\sqrt{2x+22}+4} = \frac{2}{\sqrt{16}+4} = \frac{2}{8} = 0.25

Thus, the value of the expression rounded to 2 decimal places is 0.25.

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Published 10 Oct 2026

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