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I tried to solve $$ \lim_{x\to\infty}\frac{x}{x-\sin x}. $$

After dividing by $x$ I got that it equals to: $$ \lim_{x\to\infty}\frac{1}{1-\frac{\sin x}{x}}. $$ Now, using L'hopital (0/0) I get that $$ \lim_{x\to\infty}\frac{\sin x}{x} = \lim_{x\to\infty}\cos x $$ and the lim at infinity for $\cos x$ is not defined. So basically I get that the overall limit of $$ \lim_{x\to\infty}\frac{x}{x-\sin x} $$ is $1$ or not defined?

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    $\begingroup$ You cannot use lhopitals rule for $\lim_{x\to \infty}\frac{\sin x}{x}$ $\endgroup$ – Jonathan Davidson Jul 8 '17 at 21:19
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    $\begingroup$ $(\sin x) /x \to 0$ as $x\to \infty.$ $\endgroup$ – zhw. Jul 8 '17 at 21:27
  • $\begingroup$ $$\lim_{x\to\infty}\frac{\sin(x)}x\ne\frac00$$Indeed, if it were true, that would mean$$\lim_{x\to\infty}\sin(x)=0$$which is clearly false... $\endgroup$ – Simply Beautiful Art Jul 8 '17 at 22:42
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By the squeeze theorem, we have that $$|\sin(x)|\leq 1\implies \frac{-1}{x}\leq\frac{\sin(x)}{x}\leq\frac{1}{x}\implies \lim_{x\to\infty}\frac{\sin(x)}{x}=0$$ since $\lim_\limits{x\to\infty}\frac{1}{x}=0$. You may be thinking of $\lim_\limits{x\to 0}\frac{\sin(x)}{x}=1.$

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L'Hopital is used in a wrong way. It is not the 0/0 type.

Note that $$ \lim_{x\to\infty}\frac{\sin x}{x}=0 $$ since $x\mapsto\sin x$ is bounded and $$ \lim_{x\to\infty}\frac{1}{x}=0. $$

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  • $\begingroup$ Actually, you can apply L'Hospital's rule on fractions of the form $\frac x\infty$ for any $x$, as long as the resulting limit after L'H exists (which is the problem). $\endgroup$ – Simply Beautiful Art Jul 8 '17 at 22:44
  • $\begingroup$ @Simply Beautiful Art, I have read L'H is only valid for fractions of the form $\frac00$ or $\dfrac \infty\infty$. Can you cite a source of your claim? $\endgroup$ – Mr Reality Oct 14 '17 at 11:22
  • $\begingroup$ @MrReality As mentioned on Wikipedia, you can apply L'H when the denominator goes to $\infty$, regardless of the numerator. $\endgroup$ – Simply Beautiful Art Oct 14 '17 at 11:28
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Note that $$ \left\lvert\frac{\sin x}{x}\right\rvert\le\frac{1}{x}\to0 \quad {\text{as}\quad x \to \infty}. $$ Hence $$ \lim_{x\to\infty}\frac{\sin x}{x}=0 $$ and your limit is $$ \frac{1}{1-0}=1 $$

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You have to do more work to justify applying L'Hopital's rule to pieces of expressions. The short version is that since the little limit you evaluated does not exist, the manipulations to get there are invalid. (Note that nearly every limit law says "this limit is equal to that expression containing potentially simpler limits as long as those simpler limits exist". When first learning about limits, many students let that italicized phrase fall out of their memories, but it is a very important phrase.)

Also, $\frac{\sin x}{x}$ is not a $(0/0)$ indeterminate form as $x \rightarrow \infty$. Neither $\sin x$ nor $x$ are going to zero.

Your limit can be attacked many ways. Here's one \begin{align*} \lim_{x \rightarrow \infty} \frac{x}{x- \sin x} &= \lim_{x \rightarrow \infty} \frac{x - \sin x + \sin x}{x- \sin x} \\ &= \lim_{x \rightarrow \infty} \left( \frac{x - \sin x}{x- \sin x} +\frac{\sin x}{x- \sin x} \right)\\ &= \lim_{x \rightarrow \infty} \left( 1 + \frac{\sin x}{x- \sin x} \right) \\ &= 1 + \lim_{x \rightarrow \infty} \frac{\sin x}{x- \sin x} &&\text{(if the new limit exists) using sum and constant laws} \\ &= 1 + 0 &&\text{bounded numerator, unbounded denominator} \\ &= 1 \text{.} \end{align*}

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With the sandwich theorem

$\dfrac{x}{x+1}\leq \dfrac{x}{x-\sin x}\leq \dfrac{x}{x-1} $

$\lim \limits_{x \to +\infty}\dfrac{x}{x+1}=1\quad \text{and}\quad \lim \limits_{x \to +\infty}\dfrac{x}{x-1}=1\implies \lim \limits_{x \to +\infty}\dfrac{x}{x-\sin x}=1 $

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TLDR:

$$\frac 1L=\lim_{x\to\infty}\frac{x-\sin(x)}{x}=\lim_{x\to \infty}1-\frac{\sin(x)}{x}=1$$


Explanation: Let $$f(x)=\frac{x-\sin(x)}{x}.$$ Then $$\lim_{x\to \infty}f(x)=\lim_{x\to \infty} 1-\frac{\sin(x)}{x}=1.$$ This implies the existence of an $M>0$ such that $f(x)\in (\frac 12, \frac 32)$ for every $x>M$. Since $f(x)>0$ for $x>M$, the function $g(x)=\frac{1}{f(x)}$ is well defined for $x>M$ and has limit $$\lim_{x\to \infty}\frac{x}{x-\sin(x)}=\lim_{x\to \infty} g(x)=\lim_{x\to \infty}\frac{1}{f(x)}=1$$

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  • $\begingroup$ TLDR? IMHO, this counts as a short-medium sized answer. $\endgroup$ – Simply Beautiful Art Jul 8 '17 at 22:47
  • $\begingroup$ @Simply Beautiful Art I would regard just that line as a bit Quick, at least for a beginner. The inversion $1/L$ presupposes both the existence of $L$ and the fact that $L\neq 0$. We Can show this a posteriori if $1/L=1$. $\endgroup$ – Lonidard Jul 9 '17 at 0:17

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