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PUNE MCA Exam 2017 Question Paper with answers for online practice

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Question : 27 of 41

Marks: +1, -0

\lim\limits_{x\to\infty}\frac{\sqrt{16x+1}}{\sqrt{x+2}}

is;

16
8
4
None

Solution:

Concept:

Evaluate the limit by factoring out the dominant term in both numerator and denominator.

Explanation:

Write the limit as:

\lim\limits_{x\to\infty} \frac{\sqrt{16x+1}}{\sqrt{x+2}} = \lim\limits_{x\to\infty} \frac{\sqrt{x(16+\frac{1}{x})}}{\sqrt{x(1+\frac{2}{x})}} = \lim\limits_{x\to\infty} \frac{\sqrt{x}\sqrt{16+\frac{1}{x}}}{\sqrt{x}\sqrt{1+\frac{2}{x}}}

.

Cancel

\sqrt{x}

:

\lim\limits_{x\to\infty} \frac{\sqrt{16+\frac{1}{x}}}{\sqrt{1+\frac{2}{x}}}

.

As

x\to\infty

,

\frac{1}{x}\to0

and

\frac{2}{x}\to0

, so the expression becomes

\frac{\sqrt{16}}{\sqrt{1}} = 4

.

Answer:

4

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