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For instance the $$\lim_{x=0}\frac{\sin x}{x}=1$$ Also $$\lim_{x=0}\frac{x}{\sin x}=1$$ yields the reciprocal of 1 which is 1 would this be true for all situations? Another example would be $$\lim_{x=0}\frac{1-\cos x}{x}=0$$ $$\lim_{x=0}\frac{x}{1-\cos x}=\frac{1}{0}=\text{DNE}$$

Michael Hardy
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coderhk
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    In general, if $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} g(x) = M \ne 0$, then $\lim_{x\to a} \frac{f(x)}{g(x)} = \frac{L}{M}$. If both $L$ and $M$ are zero, then the limit could exist, but more work is required. If $L\ne 0$ and $M=0$, then the limit does not exist. – Xander Henderson Sep 16 '17 at 22:37
  • There is a standard theorem that says that whenever $\lim y=L\neq 0$, then $\lim 1/y=1/L$. If $L=0$, then the limit of $1/y$ either doesn't exists, or is $\infty$, or $-\infty$. – Peter Franek Sep 16 '17 at 22:37
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    You should never write $\cdots=\text{DNE}.$ You can say something does not exist, but you should not say it is equal to does not exist. – Michael Hardy Sep 16 '17 at 22:38
  • @PeterFranek Could you please tell me the name of that theorem? I seemed to notice the relationship but I would like to see the proofs to confirm my hypothesis. – coderhk Sep 17 '17 at 01:04

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If $$\lim_{x\to a} f(x)=L,$$ then: $$\lim_{x\to a} \frac{1}{f(x)}=\frac{1}{\lim_\limits{x\to a} f(x)}=\begin{cases} \frac{1}{L}, \ \ L\ne 0, \\ \begin{cases} +\infty, L=\lim_\limits{x\to 0} x^2=0, \\ -\infty, L=\lim_\limits{x\to 0} -x^2=0, \\ undefined, L=\lim_\limits{x\to 0} x\sin{\frac{1}{x}}=0 \end{cases}, \\ undefined, \ \ f(x)=0\end{cases}$$

farruhota
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  • Don't you think that $L=0$ needs to be handled more carefully based on the sign of $f(x) $ as $x\to a$? In fact we can have the case $\infty$ or $-\infty$ and "doesn't exist". Let $a=0$ and consider $f(x) =x^{2},f(x)=-x^{2},f(x)=x$ for these three cases respectively. For the third case there is another more powerful example with $f(x) =x\sin(1/x)$. – Paramanand Singh Sep 17 '17 at 12:44
  • @ParamanandSingh, I used $\infty$ without sign, implying it can be $\pm$ depending on the function and the side of approach. For all the cases the simple answer is the limit does not exist. – farruhota Sep 17 '17 at 13:23
  • @ParamanandSingh, by your advice and examples, I specified the cases. Thank you. – farruhota Sep 17 '17 at 14:11
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In general, if $\lim_{x\to a} f(x) = L$ and $\lim_{x\to a} g(x) = M \ne 0$, then $$\lim_{x\to a} \frac{f(x)}{g(x)} = \frac{L}{M}.$$ If both $L$ and $M$ are zero, then the limit could exist, but more work is required. In the case of $\lim_{x\to 0} \frac{\sin(x)}{x}$, we can (for example) make some clever geometric arguments to show that the limit is 1. More broadly, L'Hospital's rule could allow us to compute the limit, provided that it actually applies.

Finally, if $L\ne 0$ and $M=0$, then the limit does not exist. However, it could fail to exist in two distinct ways: it could be infinite (either $\pm\infty$—for example, $\lim_{x\to 0} x^{-2} = +\infty$), or it could simply fail to exist altogether (for example, $$ \lim_{x\to 0^+} \frac{1}{x} = +\infty \qquad\text{and}\qquad \lim_{x\to 0^-} \frac{1}{x} = -\infty;$$ thus $ \lim_{x\to 0} \frac{1}{x} $ does not exit).

Xander Henderson
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