Find limit of $\lim_{x\to\infty} \frac{\ln|x|}{x}$

Question

Find limit of $\lim_{x\to\infty} \frac{\ln|x|}{x}$

I have used a substitution $\ln|x|=y$

$\lim_{y\to\infty} \frac{y}{e^y}$

=$\lim_{y\to\infty} \frac{y}{1+\frac{y}{1!}+\frac{y^2}{2!}..+\frac{y^n}{n!}}$ where n goes to infinity as well.

=$\lim_{y\to\infty} \frac{\frac{1}{y^{n-1}}}{\frac{1}{y^n}+\frac{1}{1!.y^{n-1}}+\frac{1}{2!.y^{n-2}}..+\frac{1}{n!}}$

=$\frac{0}{\frac{1}{n!}}$

=$0\times n!$ ----------(A)

=$0$

Since Maclaurin series is infinite 'n' goes to infinity. So in the line (A) can we say this multiplication is zero? If I'm wrong in any other step please correct me.

I prefer not to use L'Hospital.

Answer 1

In case you are going to use the Taylor series $e^y=1+\frac{y}{1!}+\frac{y^2}{2!}+\dots$, it is better to say that for $y>0$, $e^y=1+\frac{y}{1!}+\frac{y^2}{2!}+\dots >\frac{y^2}{2!}$, which implies that $$0<\frac{y}{e^y}< \frac{y}{\frac{y^2}{2!}}=\frac{2}{y}\to 0$$ and THEN we take the limit as $y\to +\infty$ on both sides.

Answer 2

Since the finite integral $\int_0^\infty ye^{-y}dy=1!=1$ has a non-negative integrand with unique maximum at $y=1$, the integrand's $y\to\infty$ limit is $0$.

Answer 3

$x>0$;

$\dfrac{\log x}{x}=(1/x)\displaystyle{\int_{1}^{x}}(1/t)dt \le$

$(1/x)\displaystyle{\int_{1}^{x}}(1/t^{1/2})dt =$

$ 2(1/x)(t^{1/2}) \big ]_{1}^{x}= 2(1/x)(x^{1/2}-1) $

$\lt \dfrac{2}{x^{1/2}}$.

Take the limit.

Answer 4

From your substitution, you can use the fact that the exponential function grows faster than a linear. Also,

for $x\in\mathbb R$ and $n\in\mathbb N$: $$\lim_{x\to\infty}\frac{\ln|x|}{x}=\lim_{x\to\infty}\frac{\ln|x|}{|x|}=\lim_{n\to\infty}\frac{\ln n}{n}$$ By Stolz-Cesaro for $\frac{\infty}{\infty}$ case (a discrete version of L'Hospital, i.e., you don't derive explicitly): $$\lim_{n\to\infty}\frac{\ln n}{n}=\lim_{n\to\infty}\frac{\ln (n+1)-\ln n}{(n+1)-n}=\lim_{n\to\infty}\frac{\ln\left(1+\frac{1}{n}\right)}{1}=0$$

Answer 5

Hint. Prove that for $x>0,$ we have that $$\log x<\sqrt x.$$