why the limit of this f(x) when x approach infinity is equal to infinity? why the limit of this f(x) when x approach infinity is equal to infinity?:

$$ \lim_{x\to \infty} \frac{x^2 + 4x + 5}{x-1}$$ 

$$ \lim_{x\to \infty} \frac{1 + \frac{4}{x} + \frac{5}{x^2}}{\frac{1}{x}-\frac{1}{x^2}} = \infty  $$ 

I know that all the fraction with X be denominator is equal to Zero but is that mean at the end the $ \lim_{x\to \infty} = \frac{1}{0}$ why it is equal to $ \infty $ ?
$$\lim_{x\to\infty}\left(\frac{x^2+4x+5}{x-1}\right)=$$



The leading term in the denominator of $\frac{x^2+4x+5}{x-1}$ is $x$.

So divide the numerator and denominator by this:



$$\lim_{x\to\infty}\left(\frac{x+4+\frac{5}{x}}{1-\frac{1}{x}}\right)=$$



The expressions $\frac{5}{x}$ and $\frac{1}{x}$ both tend to zero as $x$ approaches $\infty$:



$$\lim_{x\to\infty}\left(\frac{x+4+0}{1-0}\right)=$$
$$\lim_{x\to\infty}\left(\frac{x+4}{1}\right)=$$
$$\lim_{x\to\infty}\left(x+4\right)=$$
$$\lim_{x\to\infty}x=\infty$$
i dont think limit can end with 1/0 = infinity as the answer tough.

heres my approach :

instead divide it by $x^2$  divide it by $x$ instead.

so it will be $\frac{(x+4+\frac{5}{x})}{x-1} $

since $\lim _{x\to infinity} \frac{5}{x}  $ and $\lim _{x\to infinity} \frac{-1}{x}  $ are 0 . 

it leaves use with $\lim _{x\to infinity} x+4  $ which is equal to infinity