GATE CS 2021 Set 1 Q.30 (CS Q.20): Calculus

Consider the following expression.

\lim_{x \to -3} \frac{\sqrt{2x + 22} - 4}{x + 3}

The value of the above expression (rounded to 2 decimal places) is __________

Your answer

Official answer and solution

Official answer: 0.25
From: Official GATE 2021 answer key

Solution

Evaluate the limit using algebraic rationalization or L'Hopital's rule for the 0/0 indeterminate form.

Consider the limit expression:

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

Substituting $x = -3$ directly yields $\frac{\sqrt{2(-3) + 22} - 4}{-3 + 3} = \frac{\sqrt{16} - 4}{0} = \frac{0}{0}$ , which is an indeterminate form.

Method 1 (Rationalization): Multiply numerator and denominator by the conjugate $(\sqrt{2x + 22} + 4)$ :

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

L = \lim_{x \to -3} \frac{2x + 6}{(x + 3)(\sqrt{2x + 22} + 4)} = \lim_{x \to -3} \frac{2(x + 3)}{(x + 3)(\sqrt{2x + 22} + 4)}

Canceling the common $(x + 3)$ factor:

L = \frac{2}{\sqrt{2(-3) + 22} + 4} = \frac{2}{4 + 4} = \frac{2}{8} = 0.25

Method 2 (L'Hopital's Rule): Differentiating numerator and denominator with respect to $x$ :

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

Final answer

0.25

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2021 Computer Science and Information Technology (CS, Set-1)
Question: 30 of 65, printed as CS Q.20
Section: Computer Science and Information Technology

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