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 $ ?

  • $\begingroup$ $\lfloor a \rfloor >a-1$ and the interior fraction also goes to $\infty$ so the limit is $\infty$ . $\endgroup$ – user252450 Dec 3 '15 at 15:56


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

So divide the numerator and denominator by this:


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

$$\lim_{x\to\infty}\left(\frac{x+4+0}{1-0}\right)=$$ $$\lim_{x\to\infty}\left(\frac{x+4}{1}\right)=$$ $$\lim_{x\to\infty}\left(x+4\right)=$$ $$\lim_{x\to\infty}x=\infty$$

  • $\begingroup$ Thank you for your great answer, I saw in the video that it divide the top and the bottom part with x^2. Can you please also explain the process in case of divide with x^2 why it becomes infinity? $\endgroup$ – user3270418 Dec 3 '15 at 16:15
  • $\begingroup$ @user3270418 You've to look to the leading term, and that is in this case $x$ not $x^2$! $\endgroup$ – Jan Dec 3 '15 at 16:47

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


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