# How do you find the Limit of (x^2)/(lnx) as x approaches infinity?

Jun 16, 2016

It is $\infty$.

#### Explanation:

These two functions goes to infinity when $x \to \infty$ so it is in the form $\frac{\infty}{\infty}$.
Then we can appply the rule of L'Hôpital

${\lim}_{x \to \infty} \frac{{x}^{2}}{\ln} \left(x\right) = {\lim}_{x \to \infty} \frac{\frac{d}{\mathrm{dx}} {x}^{2}}{\frac{d}{\mathrm{dx}} \ln \left(x\right)}$
$= {\lim}_{x \to \infty} \frac{2 x}{\frac{1}{x}} = {\lim}_{x \to \infty} 2 {x}^{2} = \infty$

Then we discovered that ${x}^{2}$ goes to infinity "faster" than $\ln \left(x\right)$ and the ratio goes to infinity.

Jun 16, 2016

I got ${\lim}_{x \to \infty} \frac{{x}^{n}}{\ln x} = {\lim}_{x \to \infty} n {x}^{n} = \infty$.

Note that

${\lim}_{x \to \infty} \frac{{x}^{2}}{\ln x}$

is of the form $\frac{\infty}{\infty}$, since plugging in arbitrarily large numbers into ${x}^{2}$ and $\ln x$ gives a large number as well.

This fits the bill for when we can use L'Hopital's Rule (for $\frac{\infty}{\infty}$ and $\frac{0}{0}$ only!).

This rule states:

Taking the individual derivative of the numerator and denominator retains the same limit as if you didn't, if and only if the expression is of the form $\frac{\infty}{\infty}$ or $\frac{0}{0}$ and both functions are continuous in the relevant closed interval.

So, in, say, $\left[2 , \infty\right)$:

$\textcolor{g r e e n}{{\lim}_{x \to \infty} \frac{{x}^{n}}{\ln x}}$

$= {\lim}_{x \to \infty} \frac{n {x}^{n - 1}}{\frac{1}{x}}$

$= {\lim}_{x \to \infty} n {x}^{n - 1} \cdot {\left(\frac{1}{x}\right)}^{- 1}$

$= {\lim}_{x \to \infty} n {x}^{n - 1} \cdot {x}^{1}$

$= \textcolor{g r e e n}{{\lim}_{x \to \infty} n {x}^{n}}$

This turned out this way because ${x}^{n}$ accelerates towards $\infty$, whereas $\ln x$ tapers off as it approaches $\infty$. Hence, at arbitrarily large numbers, $\setminus m a t h b f \left({x}^{n}\right)$ dominates.

This means for $n = 2$, we still have:

graph{x^2/(lnx) [-5, 25, -142, 182.6]}

$\textcolor{b l u e}{{\lim}_{x \to \infty} \frac{{x}^{2}}{\ln x}}$

$= {\lim}_{x \to \infty} \frac{2 x}{\frac{1}{x}}$

$= {\lim}_{x \to \infty} 2 {x}^{2}$

$= 2 {\lim}_{x \to \infty} {x}^{2}$

$= 2 {\infty}^{2}$

$\implies \textcolor{b l u e}{\infty}$