1. ## Limits proof

The image attached to this post is my problem.

My logic is that is x^2 is in the denominator, and the limit =1, then if you just take the square root of x^2 you get x, so the limit is still 1. However, I am 99% sure I am going about this wrong.

Edit: can't change the topic name, too late now. it's a bit misleading though, sorry about that.

2. Originally Posted by xezial
The image attached to this post is my problem.

My logic is that is x^2 is in the denominator, and the limit =1, then if you just take the square root of x^2 you get x, so the limit is still 1. However, I am 99% sure I am going about this wrong.

Edit: can't change the topic name, too late now. it's a bit misleading though, sorry about that.
$\displaystyle \lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{xf(x) }{x^2}=\left(\lim_{x\to0}x\right)\left(\lim_{x\to0 }\frac{f(x)}{x^2}\right)=(0)(1)=0$

3. Originally Posted by xezial
The image attached to this post is my problem.

My logic is that is x^2 is in the denominator, and the limit =1, then if you just take the square root of x^2 you get x, so the limit is still 1. However, I am 99% sure I am going about this wrong.

Edit: can't change the topic name, too late now. it's a bit misleading though, sorry about that.
We are given:

$\displaystyle \lim_{x \to 0} \frac{f(x)}{x^2}=1$

So for any $\displaystyle \varepsilon>0$, for $\displaystyle x$ small enough:

$\displaystyle 1-\varepsilon<\frac{f(x)}{x^2}<1+\varepsilon$

or:

$\displaystyle (1-\varepsilon)x<\frac{f(x)}{x}<(1+\varepsilon)x$

hence by the squeeze theorem $\displaystyle \lim_{x \to 0} \frac{f(x)}{x}=??$

CB

4. Wow, I haven't really gotten that far into calc to be familiar with the squeeze theorem and all that... but is the solution redsox fan posted legitamate?

5. Originally Posted by xezial
Wow, I haven't really gotten that far into calc to be familiar with the squeeze theorem and all that... but is the solution redsox fan posted legitamate?
Yes.