Maliyah Robles

Answered

2022-07-04

I have to calculate the following limit

$\underset{x\to +\mathrm{\infty}}{lim}\frac{x\mathrm{ln}x}{1-\mathrm{sin}x}$

$\underset{x\to +\mathrm{\infty}}{lim}\frac{x\mathrm{ln}x}{1-\mathrm{sin}x}$

Answer & Explanation

Dalton Lester

Expert

2022-07-05Added 12 answers

$\underset{n\to \mathrm{\infty}}{lim}\frac{n}{2+\pm (-1{)}^{n}}$

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 $\underset{x\to \mathrm{\infty}}{lim}f(x).$

In your case, you easily get, where f is defined, and $x>1,$

$f(x)=\frac{x\mathrm{ln}x}{1-\mathrm{sin}x}\ge \frac{x\mathrm{ln}x}{2},$

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 $f(x)\to \mathrm{\infty}.$

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 $\underset{x\to \mathrm{\infty}}{lim}f(x).$

In your case, you easily get, where f is defined, and $x>1,$

$f(x)=\frac{x\mathrm{ln}x}{1-\mathrm{sin}x}\ge \frac{x\mathrm{ln}x}{2},$

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 $f(x)\to \mathrm{\infty}.$

racodelitusmn

Expert

2022-07-06Added 5 answers

The quotient $\frac{x\mathrm{ln}(x)}{1-\mathrm{sin}(x)}$ is undefined when $x\in {\displaystyle \frac{\pi}{2}}+2\pi \mathbb{Z}$. Otherwise, $1-\mathrm{sin}(x)\u2a7d2$, and therefore

$\begin{array}{}\text{(1)}& \frac{x\mathrm{ln}(x)}{1-\mathrm{sin}(x)}\u2a7e\frac{x\mathrm{ln}(x)}{2}.\end{array}$

Since

$\underset{x\to \mathrm{\infty}}{lim}x\mathrm{ln}(x)=\mathrm{\infty},$

it follows from (1) that

$\underset{x\to \mathrm{\infty}}{lim}\frac{x\mathrm{ln}(x)}{1-\mathrm{sin}(x)}=\mathrm{\infty}$

too.

$\begin{array}{}\text{(1)}& \frac{x\mathrm{ln}(x)}{1-\mathrm{sin}(x)}\u2a7e\frac{x\mathrm{ln}(x)}{2}.\end{array}$

Since

$\underset{x\to \mathrm{\infty}}{lim}x\mathrm{ln}(x)=\mathrm{\infty},$

it follows from (1) that

$\underset{x\to \mathrm{\infty}}{lim}\frac{x\mathrm{ln}(x)}{1-\mathrm{sin}(x)}=\mathrm{\infty}$

too.

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