lim _{x rightarrow 0} frac{48}{x^4} int _{0}^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given limit is $L = \lim _{x \rightarrow 0} \frac{48}{x^4} \int _{0}^{x} \frac{t^3}{t^6+1} dt$. This is a $\frac{0}{0}$ indeterminate form. Applyi

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) Given limit is $L = \lim _{x ightarrow 0} \frac{48}{x^4} \int _{0}^{x} \frac{t^3}{t^6+1} dt$. This is a $\frac{0}{0}$ indeterminate form. Applying $L'	ext{Hospital's Rule}$,we differentiate the numerator and denominator with respect to $x$: $L = 48 \lim _{x ightarrow 0} \frac{\frac{d}{dx} \int _{0}^{x} \frac{t^3}{t^6+1} dt}{\frac{d}{dx} (x^4)}$. Using the $	ext{Leibniz Integral Rule}$,$\frac{d}{dx} \int _{0}^{x} f(t) dt = f(x)$: $L = 48 \lim _{x ightarrow 0} \frac{\frac{x^3}{x^6+1}}{4x^3}$. Simplifying the expression: $L = 48 \lim _{x ightarrow 0} \frac{x^3}{4x^3(x^6+1)} = 48 \lim _{x ightarrow 0} \frac{1}{4(x^6+1)}$. Evaluating the limit as $x ightarrow 0$: $L = \frac{48}{4(0^6+1)} = \frac{48}{4} = 12$.

$\lim _{x \rightarrow 0} \frac{48}{x^4} \int _{0}^{x} \frac{t^3}{t^6+1} dt$ is equal to $.......$.

Similar Questions

$\int_0^{\pi /2} \sin^2 x \cos^3 x \, dx = $

The value of $\int_{-3\pi}^{3\pi} \sin^2 \theta \sin^2 2\theta \, d\theta$ is equal to:

If $f(x) = \int_{0}^{x} t \sin t \,dt,$ then $f^{\prime}(x)$ is

The number of solutions of the equation $\frac{d}{dx} \int_{\cos x}^{\sin x} \frac{dt}{\sqrt{1 - t^2}} = 2\sqrt{2}$ in the interval $[0, \pi]$ is:

$\mathop {Lim}\limits_{k \to 0} \frac{1}{k} \int\limits_0^k (1 + \sin 2x)^{\frac{1}{x}} dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module