Maliyah Robles

2022-07-04

I have to calculate the following limit
$\underset{x\to +\mathrm{\infty }}{lim}\frac{x\mathrm{ln}x}{1-\mathrm{sin}x}$

Dalton Lester

Expert

$\underset{n\to \mathrm{\infty }}{lim}\frac{n}{2+±\left(-1{\right)}^{n}}$
is infinity even though the denominator does not converge.
The real problem with your function is that it isn’t defined for all of $\mathbb{R}.$ But if f is a function defined on a subset of $\mathbb{R}$ with no real upper bound, we can still define $\underset{x\to \mathrm{\infty }}{lim}f\left(x\right).$
In your case, you easily get, where f is defined, and $x>1,$
$f\left(x\right)=\frac{x\mathrm{ln}x}{1-\mathrm{sin}x}\ge \frac{x\mathrm{ln}x}{2},$
since $0<1-\mathrm{sin}x\le 2$ in the domain of f
Since $\frac{x\mathrm{ln}x}{2}\to \mathrm{\infty },$ this means $f\left(x\right)\to \mathrm{\infty }.$

racodelitusmn

Expert

The quotient $\frac{x\mathrm{ln}\left(x\right)}{1-\mathrm{sin}\left(x\right)}$ is undefined when $x\in \frac{\pi }{2}+2\pi \mathbb{Z}$. Otherwise, $1-\mathrm{sin}\left(x\right)⩽2$, and therefore
$\begin{array}{}\text{(1)}& \frac{x\mathrm{ln}\left(x\right)}{1-\mathrm{sin}\left(x\right)}⩾\frac{x\mathrm{ln}\left(x\right)}{2}.\end{array}$
Since
$\underset{x\to \mathrm{\infty }}{lim}x\mathrm{ln}\left(x\right)=\mathrm{\infty },$
it follows from (1) that
$\underset{x\to \mathrm{\infty }}{lim}\frac{x\mathrm{ln}\left(x\right)}{1-\mathrm{sin}\left(x\right)}=\mathrm{\infty }$
too.

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