Question

13) Utilizando a regra de L’Hopital, determine os seguintes limites:

a)$\lim_{X \to1} \frac{X^1^0^0-X^2+X-1}{x^1^0-1}$

b)$\lim_{X \to \infty +1} e^3^x/x^2$

c)$\lim_{X \to 0} e^x-e^-^x/seno (x)$

d)$\lim_{x \to + \infty} log(x)/x^3$

e)$\lim_{x \to + \infty} \ x^2e^-^x$

f)$\lim_{x \to 0+\ xe^ \frac{1}{x}$

Answer 1

$a) \lim_{x \to \01} \frac{x^{100}-x^{2}+x-1}{x^{10}-1} = \frac{100x^{99}-2x+1}{10x^{9}} = \frac{100-2+1}{10} = \frac{99}{10}$

$b) \lim_{x\to \infty} \frac{e^{3x}}{x^{2}} = \frac{3e^{3x}}{2x} = \frac{9e^{3x}}{2} = \frac{{\infty}}{2} = {\infty}$

$c) \lim_{x \to \00} \frac{e^x-e^{-x}}{sen(x)} = \frac{e^{x}+e^{-x}}{cos(x)} = \frac{1+1}{1} = 2$

$d) \lim_{n \to \infty} \frac{log(x)}{x^3} = \frac{ \frac{1}{x}log(e) }{3x^2} = \frac{log(e)}{3x^3} = 0$

$e) \lim_{x \to \infty} x^2e^{-x} = \frac{x^2}{e^x} = \frac{2x}{e^x} = \frac{2}{e^x} = 0$

$f) \lim_{x \to \00+} xe^{ \frac{1}{x}} = e^{ \frac{1}{x}}-\frac{e \frac{1}{x} }{x} } = \frac{e^{ \frac{1}{x}}(x-1) }{x} = \frac{ {\infty}(-1)}{0+} = \frac{- {\infty}}{0+} = - {\infty}$