1
$\begingroup$

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

  • 0
    It is good practice to make the post self-contained, see http://meta.math.stackexchange.com/questions/2483/broken-imgur-links http://meta.math.stackexchange.com/questions/2674/how-self-contained-should-questions-be and other topics in meta linked to these ones. I've edited your post, you should check whether some further edits are needed.2011-10-16

2 Answers 2

6

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 $\lim\limits_{x\to0} x^2$?

6

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$