lim limits_{x rightarrow 1} left( frac{int li | vedclass

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

Question

(D) Let $L = \lim \limits_{x \rightarrow 1} \frac{\int_{0}^{(x-1)^{2}} t \cos(t^{2}) dt}{(x-1) \sin(x-1)}$.This is a $\frac{0}{0}$ form.Using the Leib

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(D) Let $L = \lim \limits_{x \rightarrow 1} \frac{\int_{0}^{(x-1)^{2}} t \cos(t^{2}) dt}{(x-1) \sin(x-1)}$.This is a $\frac{0}{0}$ form.Using the Leibniz rule for differentiation under the integral sign,the derivative of the numerator is $\frac{d}{dx} \int_{0}^{(x-1)^{2}} t \cos(t^{2}) dt = (x-1)^{2} \cos((x-1)^{4}) \cdot \frac{d}{dx}((x-1)^{2}) = (x-1)^{2} \cos((x-1)^{4}) \cdot 2(x-1) = 2(x-1)^{3} \cos((x-1)^{4})$.The derivative of the denominator is $\frac{d}{dx} ((x-1) \sin(x-1)) = \sin(x-1) + (x-1) \cos(x-1)$.Applying $L$'Hopital's rule:$L = \lim \limits_{x \rightarrow 1} \frac{2(x-1)^{3} \cos((x-1)^{4})}{\sin(x-1) + (x-1) \cos(x-1)}$.Divide numerator and denominator by $(x-1)$:$L = \lim \limits_{x \rightarrow 1} \frac{2(x-1)^{2} \cos((x-1)^{4})}{\frac{\sin(x-1)}{x-1} + \cos(x-1)}$.As $x \rightarrow 1$,$(x-1) \rightarrow 0$,so $\frac{\sin(x-1)}{x-1} \rightarrow 1$ and $\cos(x-1) \rightarrow 1$.$L = \frac{2(0)^{2} \cdot \cos(0)}{1 + 1} = \frac{0}{2} = 0$.

$\lim \limits_{x \rightarrow 1} \left( \frac{\int \limits_{0}^{(x-1)^{2}} t \cos(t^{2}) dt}{(x-1) \sin(x-1)} \right)$ is equal to

Similar Questions

$\int_0^\pi \sin^5\left( \frac{x}{2} \right) \, dx$ equals

Let $f(x) = \int_{\sin x}^{\cos x} e^{-t^2} dt$. Then $f^{\prime}\left(\frac{\pi}{4}\right)$ equals

Let $f(x) = \int\limits_0^{x^2} {(t - 1)(t - 4)(t - 9)} dt$,then:

$\int_{-2}^2 x^4(4-x^2)^{\frac{7}{2}} dx=$

If $\int_0^x {f(t)\,dt} = x + \int_x^1 {t\,f(t)\,dt,}$ then the value of $f(1)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module