mathop {lim }limits_{x to infty } frac{{{x^n} | vedclass

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

Question

(B) Using $L$'$H$ôpital's Rule repeatedly,we differentiate the numerator and denominator $n$ times:$\mathop {\lim }\limits_{x 	o \infty } \frac{{{x^n

Vedclass · Accepted Answer

$n$ is any whole number

Vedclass · Answer

(B) Using $L$'$H$ôpital's Rule repeatedly,we differentiate the numerator and denominator $n$ times:$\mathop {\lim }\limits_{x 	o \infty } \frac{{{x^n}}}{{{e^x}}} = \mathop {\lim }\limits_{x 	o \infty } \frac{{n \cdot x^{n-1}}}{{{e^x}}} = \mathop {\lim }\limits_{x 	o \infty } \frac{{n(n-1) \cdot x^{n-2}}}{{{e^x}}} = \dots = \mathop {\lim }\limits_{x 	o \infty } \frac{{n!}}{{{e^x}}}$.Since $\mathop {\lim }\limits_{x 	o \infty } e^x = \infty$,we have $\frac{n!}{\infty} = 0$.This holds true for any whole number $n \ge 0$.

$\mathop {\lim }\limits_{x \to \infty } \frac{{{x^n}}}{{{e^x}}} = 0$ for

Similar Questions

If $a > 0$ and $\lim _{x \rightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$,then $a$ is equal to

If $f(x)$ is a differentiable function of $x$,then $\mathop {Limit}\limits_{h \to 0} \frac{f(x + 3h) - f(x - 2h)}{h} = $

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{8 \sqrt{2}-(\cos x+\sin x)^{7}}{\sqrt{2}-\sqrt{2} \sin 2 x}$ is equal to

Evaluate the limit: $\mathop {\lim }\limits_{x \to 0} \frac{{\int\limits_0^x (\tan^{-1} t)^2 dt}}{{\sin x - x}}$

$\mathop {\lim }\limits_{x \to 0} \frac{{\ln (\cos x)}}{{{x^2}}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module