Question

Find \(lt_{x\rightarrow 0}\frac{(3e^x-2e^{2x}-e^{3x})}{(e^x+e^{2x}-2e^{3x})}\)

(a) ^3⁄2

(b) 0

(c) ^4⁄3

(d) –^4⁄3

I have been asked this question in an online interview.

This interesting question is from Indeterminate Forms topic in portion Differential Calculus of Engineering Mathematics

Answer 1

Right option is (c) ^4⁄3

The explanation: Form is  0 / 0

Applying L hospitals rule we have

\(=lt_{x\rightarrow 0}\frac{3e^x-4e^{2x}-3e^{3x}}{e^x+2e^{2x}-6e^{3x}}\)

\(=\frac{3-4-3}{1+2-6}\)

\(=\frac{4}{3}\)