Question

Evaluate \(\lim_{x\rightarrow\infty}\left [\frac{x-1}{x-2} \right ]^x\).

(a) ^1⁄4

(b) ^1⁄3

(c) ^1⁄2

(d) 1

This question was addressed to me in semester exam.

The doubt is from Indeterminate Forms topic in division Differential Calculus of Engineering Mathematics

Answer 1

Right choice is (c) ^1⁄2

The explanation is: \(y=\lim_{x\rightarrow\infty}\left [\frac{x-1}{x-2} \right ]^x\)

\(ln y=\lim_{x\rightarrow\infty}xln\left [\frac{x-1}{x-2} \right ]\)

=>\(\lim_{x\rightarrow\infty}\frac{\left [\frac{x-1}{x-2} \right ]}{\frac{1}{x}}\)

By putting 1=1/y, we get

=>\(\lim_{y\rightarrow 0}\frac{ln\left [\frac{x-1}{x-2} \right ]}{y}=[ln(1/2)]/0\) (i.e indeterminate)

Hence by applying L’Hospital rule

=>\(\lim_{y\rightarrow 0}\frac{ln\left [\frac{x-1}{x-2} \right ]}{y}=\lim_{y\rightarrow 0}\frac{\frac{2-y-1+y}{(2-y)^2}}{\frac{1-y}{2-y}}=\lim_{y\rightarrow 0}\frac{\frac{1}{2-y}}{\frac{1-y}{1}}=\lim_{y\rightarrow 0}(\frac{1}{(2-y)(1-y)})\)=1/2