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Interchange log and limit: TIFR GS 2017 Part A Problem 4

Understand the problem

True or False?
Over the real line,
\( \lim_{x\to \infty} log(1+ \sqrt{4+x} – \sqrt{1+x)}= log 2 \).

Source of the problem
TIFR GS 2017 MATHEMATICS ENTRANCE EXAMINATION PAPER
Topic
Real analysis
Difficulty Level
easy
Suggested Book
S.K. Mapa and Bertle and Sherbert.

Start with hints

Do you really need a hint? Try it first!

Since \(log\) is a continuous function, \( \lim_{x\to \infty} log(1+ \sqrt{4+x} – \sqrt{1+x}) = log \lim_{x\to \infty}(1+ \sqrt{4+x} – \sqrt{1+x})\). Rationalize the portion which is in the square root.
\(\log \lim_{x\to \infty}(1+ \frac{\sqrt{4+x} -\sqrt{1+x}}{\sqrt{4+x} + \sqrt{1+x}} ( \frac{\sqrt{4+x} + \sqrt{1+x}}{ \sqrt{4+x}+ \sqrt{1+x}})) \) = \(\log \lim_{x\to \infty}(1 + \frac {3}{(\sqrt{4+x}+ \sqrt{1+x})^2}) \)
Calculate the limit. You will get the value \( \log1= 0\). But the given value of the limit is \( log2 \) Hence, the statement is false.

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