Question

$\lim _ { x \rightarrow \infty } \frac { e ^ { x } } { x ^ { 2 } + 1 }$ ​

Answer 1

Resposta:

lim    e^(x)  / (x²+1)  =∞/∞

x-->∞

Utilizando L'Hopital  ...derivando em cima e embaixo

lim   e^(x)/2x

x-->∞

..novamente

lim   e^(x)/2   =   ∞   é a resposta

x-->∞

Answer 2

$\displaystyle \lim_{\text x\to \infty} \frac{\text e^{\text x}}{\text x^2+1} \\\\\\ \text{ao fazer}\ \text x\to \infty\ \text{vamos ter indetermina{\c c}{\~a}o do tipo}: \frac{\infty}{\infty} \\\\ \underline{\text{Ent{\~a}o vamos usar a regra de L'hospital}}:$

$\displaystyle \lim_{\text x\to \infty} \frac{\text e^{\text x}}{\text x^2+1}= \lim_{\text x\to \infty} \frac{[\text e^{\text x}]'}{[\text x^2+1]'} = \lim_{\text x\to \infty} \frac{\text e^{\text x}}{2\text x} \\\\\\ \text{Ainda h{\'a} indetermina{\c c}{\~a}o, ent{\~a}o} : \\\\ \lim_{\text x\to\infty} \frac{[\text e^{\text x}]'}{[2\text x]'} = \lim_{\text x\to\infty} \frac{\text e^{\text x}}{2 } \to \frac{1}{2}.\lim_{\text x\to \infty } \text e^{\text x}$

$\displaystyle \frac{1}{2}.\lim_{\text x\to \infty } \text e^{\text x} = \frac{1}{2}.\text e^{\infty } = \frac{1}{2}.\infty = \infty \\\\\\ \underline{\text{Portanto}}: \\\\ \huge\boxed{\lim_{\text x\to\infty} \frac{\text e^{\text x}}{\text x^2+1} = \infty \ }\checkmark$