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I am reviewing my book but it has no answers, just if the statement is true or false with no justification. if $$\lim_{x\to5}f(x)=0$$ and $$\lim_{x\to5}g(x)=0$$ then $$\lim_{x\to5}f(x)/g(x)$$ does not exist

I said true but the book says false and I do not see why, 0/0 is undefined.

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Imagine something like $f(x) = g(x) = x - 5$. The theorems about limits tell you something about a limit of a quotient only if the limit of the denominator is non-zero. – Dylan Moreland May 4 '12 at 23:24
@DylanMoreland Oh I see that now, thanks. – user138246 May 4 '12 at 23:26
If this was true, L'Hopital's rule would not be quite as useful. – AMPerrine May 4 '12 at 23:35
up vote 2 down vote accepted

When in doubt you can sometimes try to consider easy toy examples. For instance, if you let $f(x) = x -5$ and $g(x) = x - 5$, you have

$$\lim_{x \rightarrow 5} \frac{x - 5}{x - 5} = 1$$

so in this case the limit does indeed exist. Now if instead you have $f(x) = x- 5$ but $g(x) = (x - 5)^2$ then

$$\lim_{x \rightarrow 5} \frac{x - 5}{(x - 5)^2} = \lim_{x \rightarrow 5} \frac{1}{x - 5} = \pm \infty$$

depending on whether you approach $5$ from the left or from the right, but in this case the limit does not exist.

So this is basically the reason why an expression with limits in which you have $0/0$ is undefined. Sometimes you can get a limit that exists like in the first case where the answer is $1$, but other times you can have a limit that doesn't even exist.

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Limit of sin x / x would be an example where the limit can exist for a case of 0/0.

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