Question

$\underset{x\to 0}{\lim }\frac{(1-e^x)\sin x}{(x^2+x^3)}$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A

$-1$

Explanation for the correct option:

Finding the value of the given function after applying the limits:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{(1-e^x)\sin x}{(x^2+x^3)} \\ =\underset{x\to 0}{\lim }\frac{(1-e^x)\sin x}{(x^2+x^3)} \\ =\underset{x\to 0}{\lim }\frac{(1-e^x)\sin x}{x^2\left(1+x\right)}\left[(x^2+x^3)\mathrm{and}\mathrm{be}\mathrm{written}\mathrm{as}{x}^2\left(1+x\right)\right] \\ =\underset{x\to 0}{\lim }\frac{1-e^x}{x}\times \frac{\sin x}{x}\times \frac{1}{1+x}\mathbf{}\left[\mathrm{Splitting}\mathrm{the}\mathrm{terms}\right] \\ =\underset{x\to 0}{\lim }\frac{-(e^x-1)}{x}\times \frac{\sin x}{x}\times \frac{1}{1+x} \end{gathered}$$

Applying the limits,

$$\begin{gathered} =\left(-1\right)\times \left(1\right)\times \left(\frac{1}{1}\right)\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{[}\underset{\mathbf{\theta }\mathbf{\to }\mathbf{0}}{\mathbf{lim}}\frac{\mathbf{sin}\left(\theta \right)}{\mathbf{\theta }}\mathbf{=}\mathbf{1}\mathbf{]} \\ =-1 \end{gathered}$$

Therefore, the correct answer is option (A).