True or False?

Over the real line,

\( \lim_{x\to \infty} log(1+ \sqrt{4+x} – \sqrt{1+x)}= log 2 \).

TIFR GS 2017 MATHEMATICS ENTRANCE EXAMINATION PAPER

Real analysis

easy

S.K. Mapa and Bertle and Sherbert.

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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