Question

Find the value of \(\lim_{x\rightarrow 0}\frac{xsin(x-1)}{(x-1)e^x}\) is, where {x} is the fractional part of x.

(a) ^2⁄e

(b) ^1⁄e

(c) 0

(d) –^1⁄e

The question was asked during an interview.

I want to ask this question from Indeterminate Forms in chapter Differential Calculus of Engineering Mathematics

Answer 1

Right option is (b) ^1⁄e

Best explanation: \(\lim_{{x}\rightarrow 1^+}\frac{{x}sin(x-1)}{(x-1)e^x}\)

\(=\lim_{x\rightarrow 1^+}{x}\lim_{x\rightarrow 1^+}{\frac{sin(x-1)}{x-1}} \lim_{x\rightarrow 1^+}\frac{1}{e^x}=\lim_{x\rightarrow 1^+}{x}1.\frac{1}{e}=\frac{1}{e}\)

because, \(\lim_{x\rightarrow 1^+}{x}=1\).