Question

$\underset{x\to 0}{\lim }\frac{8\sin x+x\cos x}{3\tan x+x^2}=$

• A. $3$
• B. $2$
• C. $-1$
• D. $4$

Answer 1

The correct option is A

$3$

Explanation for the correct option:

Expanding the given equation and applying the limits:

$$\begin{gathered} \underset{x\to 0}{\lim }\frac{8\sin x+x\cos x}{3\tan x+x^2} \\ \mathrm{Applying}\mathrm{L}\text{'}\mathrm{hospital}\text{'}\mathrm{s}\mathrm{Rule}\left[\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)\mathbf{=}\underset{\mathbf{x}\mathbf{\to }\mathbf{c}}{\mathbf{l}\mathbf{i}\mathbf{m}}\left(\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}\right)\right] \\ =\underset{x\to 0}{\lim }\frac{8\sin x+\cos x+x(-\sin x)}{3{\sec }^{2}x+2x}\left[\frac{\mathbf{d}\mathbf{\left(}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{x}\mathbf{\right)}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\frac{\mathbf{d}\mathbf{(}\mathbf{3}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{x}}^{\mathbf{2}}\mathbf{)}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathbf{3}\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{x}\mathbf{+}\mathbf{2}\mathit{x}\right] \end{gathered}$$

Applying the limits,

$$\begin{gathered} =\frac{8\sin 0+\cos 0+0(-\sin 0)}{3{\sec }^{2}0+2\left(0\right)} \\ =\frac{8+1+0}{3\left(1\right)+0} \\ =\frac{9}{3} \\ =3 \end{gathered}$$

Thus, $\underset{x\to 0}{\lim }\frac{8\sin x+x\cos x}{3\tan x+x^2}=3$

Therefore, the correct answer is option (A).