$\begingroup$What is the difference (if any) between-
$$\lim_{h\to0}\frac{0}{h} \text{ and } \lim_{h\to\infty}\frac{0}{h}$$
I argue that both must be $=0$ since the numerator is exactly $0$. But one fellow refuses to agree and argues that the first limit can't be $0$ as anything finite by something tending to $0$ is always $\infty$.
So,how can I explain it to the person? Also, if possible can anyone provide some good reference on this particular issue (Apostol perhaps)?
Thanks for any help!
$\endgroup$63 Answers
$\begingroup$In both cases you are dividing zero by a nonzero number .
Thus your fraction is identically zero and as a result the limit is zero.
$\endgroup$$\begingroup$For the first one let me make an intuitive explanation of what's going on with $\lim_{h\to 0}\frac{0}{h}$.
The above limit means "What value does the above expression get while $h$ gets arbitrarily close to $0$", meaning that $h$ is number very close to $0$ but not equal to that. Suppose $h=0.000001$. Then $\frac{0}{h}=\frac{0}{0.000001}=0$. Even if this number gets even closer to $0$ you see that the expression continuous to be equal to $0$. That's because $0$ divided by any number obsiously gives $0$ as the answer. Hence $\lim_{h\to 0}\frac{0}{h}=0$.
I wish I helped!
$\endgroup$$\begingroup$When in doubt take it back to the definition.
$\lim_\limits{x\to 0} f(x) = 0$ means
$\forall \epsilon >0, \exists \delta >0 : 0<|x|<\delta \implies |f(x)|<\epsilon$
As $f(x) = 0$ for all $x \ne 0$ the definition above is satisfied.
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