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I just can't figure out this problem. I would like to know how I can figure it out more than just the answer. The x's need to be canceled out to figure it out but I can't think of how to do that without an equation.


If $\lim\limits_{x\to0} \frac{f(x)}{x^2} = 9$, evaluate the following limits:

  • (a) $\lim\limits_{x\to0} f(x)$.

  • (b) $\lim\limits_{x\to0} \frac{f(x)}x$

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    $\begingroup$ What does the limit in the question mean? Essentially that $f(x) \sim 9x^2$ around $0$. From that, you can guess what the two other limits are. The rest is just a matter of formalizing the answers, if you need a proof. $\endgroup$ – Raskolnikov Oct 16 '11 at 8:18
  • $\begingroup$ Hint: the actual limit is meaningless, it is enough to know that the limit exists. $\endgroup$ – yaakov Oct 16 '11 at 8:19
  • $\begingroup$ It is good practice to make the post self-contained, see meta.math.stackexchange.com/questions/2483/broken-imgur-links meta.math.stackexchange.com/questions/2674/… and other topics in meta linked to these ones. I've edited your post, you should check whether some further edits are needed. $\endgroup$ – Martin Sleziak Oct 16 '11 at 8:24
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Hint (or maybe even half of the solution, I should say): $$f(x)=\frac{f(x)}{x^2}\cdot x^2$$ What can you say about the limit of $f(x)$ if you know the values of $\lim\limits_{x\to0} \frac{f(x)}{x^2}$ and $\lim\limits_{x\to0} x^2$?

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Just think of the limits as:

$$ \lim_{x \to 0} f(x)=\lim_{x\to 0} \ x^2 \cdot \left( \frac{f(x)}{x^2}\right)=0 \cdot 9$$

and

$$ \lim_{x \to 0} \frac{f(x)}{x}=\lim_{x \to 0} \ x \cdot \left( \frac{f(x)}{x^2}\right)=0\cdot 9$$

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