_ x 0 [x (1+3 x) (e^ 3 x -1 )^2 ]= - Vidyadip

MCQ

$\lim _{x \rightarrow 0}\left[\frac{x \cdot \log (1+3 x)}{\left(e^{3 x}-1\right)^2}\right]=$

A
$\frac{1}{e^9}$
B
$\frac{1}{\mathrm{e}^3}$
C
$\frac{1}{9}$
✓
$\frac{1}{3}$

✓

Answer

Correct option: D.

$\frac{1}{3}$

(D) $\frac{1}{3}$
Hint:
$\lim _{x \rightarrow 0} \frac{x \cdot \log (1+3 x)}{\left(e^{3 x}-1\right)^2}$
$=\frac{\lim _{x \rightarrow 0} \frac{\log (1+3 x)}{x}}{\lim _{x \rightarrow 0}\left(\frac{e^{3 x}-1}{x}\right)^2} $
$=\frac{\lim _{x \rightarrow 0}\left[\frac{\log (1+3 x)}{3 x} \times 3\right]}{\lim _{x \rightarrow 0}\left[\left(\frac{e^{3 x}-1}{3 x}\right)^2 \times(3)^2\right]}=\frac{1}{3}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Limits→Limits — all question types→Maths STD 11 Science question papers→Maharashtra Board STD 11 Science question papers→Maharashtra Board question paper generator→

Similar questions

If $z_1=\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$ and $z_2=\sqrt{3}\left(\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}\right)$, then $\left|z_1 z_2\right|$ is

If 5x - 12y + 10 = 0 and 12y - 5x + 16 = 0 are tangents of a circle, then radius of that circle is

The value of $9^{\frac13}.9^{\frac19}.9^{\frac{1}{27}}\dots\text{to }\infty,$ is :

Equation of the parabola with vertex at the origin and directrix with equation x + 8 = 0 is ________

The point on the curve $y^2=x$, the tangent at which makes an angle $45^{\circ}$ with $X$-axis will be

The value of the expression $\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \ldots \cos 179^{\circ}=$

A single letter is selected from the word TRICKS. The probability that it is either T or R is

If the point $(5, 2)$ bisects the intercept of a line between the axes, then its equation is :

If $f(x)=\frac{e^{x^2}-\cos x}{x^2}$, for $x \neq 0$ is continuous at $x=0$, then value of $f(0)$ is

Given the function $f (x)=\frac{ a + a }{2}, a > 2$, then