# How do you find the limit of lnx/(sqrtx+lnx) as x->oo?

Nov 6, 2016

${\lim}_{x \rightarrow \infty} \ln \frac{x}{\sqrt{x} + \ln x} = 0$

#### Explanation:

$L = {\lim}_{x \rightarrow \infty} \ln \frac{x}{\sqrt{x} + \ln x}$

This is an indeterminate form, in the form of $\frac{\infty}{\infty}$. Due to that, we can use L'Hopital's rule by differentiating the numerator and denominator individually.

$L = {\lim}_{x \rightarrow \infty} \frac{\frac{d}{\mathrm{dx}} \ln x}{\frac{d}{\mathrm{dx}} \left(\sqrt{x} + \ln x\right)} = {\lim}_{x \rightarrow \infty} \frac{\frac{1}{x}}{\frac{1}{2 \sqrt{x}} + \frac{1}{x}}$

Multiply through by $\frac{x}{x}$:

$L = {\lim}_{x \rightarrow \infty} \frac{\frac{1}{x} \cdot x}{\left(\frac{1}{2 \sqrt{x}} + \frac{1}{x}\right) \cdot x} = {\lim}_{x \rightarrow \infty} \frac{1}{\frac{\sqrt{x}}{2} + 1}$

As we examine this, we see the denominator approaches infinity as the numerator stays unchanged, so the overall fraction will approach $0$.

$L = 0$

Another way of approaching this is that both the numerator and denominator of the fraction contain $\ln x$, so those will increase at the same rate. However, the denominator is also increasing with $\sqrt{x}$, so the denominator will be growing faster than the numerator thanks to the additional $\sqrt{x}$. Thus, the limit will be $0$ because the denominator will constantly be outgrowing the numerator.