# Math Help - Finding the limit of a particular trig function as x approaches 0

1. ## Finding the limit of a particular trig function as x approaches 0

Consider the problem, $\lim_{x\to\0}\frac{sinx}{2x^2-x}=-1$.

I tried u-substitution; let $u=2x^2-x$, then $\lim_{x\to\0}u=0$;

for, $(u=2x^2-x)$ $\vee$ $(\lim_{x\to\0}2x^2-x=\lim_{x\to\0}2(0)^2-0=0)$ $\subset$ $\lim_{x\to\0}u=0$.

From this,

$\lim_{x\to\0}\frac{sinx}{2x^2-x}=\frac{\lim_{x\to\0}sinx}{\lim_{u\to\0}u}$ is effectively, $\lim_{x\to\0}\frac{sinx}{x}=1$

I figured I could treat the limits for the numerator and denominator independently as it doesn't seem to violate any limit laws—is this okay?

The question: where did the negative sign arise from?

2. ## Re: Finding the limit of a particular trig function as x approaches 0

Originally Posted by Lambin
Consider the problem, $\lim_{x\to\0}\frac{sinx}{2x^2-x}=-1$.
I wish I knew what dummy mathematics-educator came up with u-substitution.
It has caused more damage than any other concept that I see.
My advise is to use that concept as a last resort.

Look: $\frac{sinx}{2x^2-x}=\left(\frac{\sin(x)}{x}\right)\left(\frac{1}{2x-1}\right)$.

Now if you cannot simply look at that and see the limit, then you are not ready for this material.

If you are not able to do that, go back a more elementary course. It will save yourself many tears.

3. ## Re: Finding the limit of a particular trig function as x approaches 0

Originally Posted by Lambin

I figured I could treat the limits for the numerator and denominator independently as it doesn't seem to violate any limit laws—is this okay?

The question: where did the negative sign arise from?
Just because it doesn't clearly violate any laws doesn't mean its an okay step. Even if you did that substitution, which is okay in itself by the way, you need to apply the same substitution to the numerator as well. This would require finding x as a function of u, a long and tedious way to solve this problem. Plato's way is the most direct one.

The reason your logic at the last step is incorrect, by the way, is that $\frac{\lim_{x\to\0}sinx}{\lim_{u\to\0}u} = \frac{0}{0}$, and so you are back to the indeterminate form. You can't arbitrarily change the variable in the denominator back to x just because it would be convenient - x and u do not approach 0 at the same pace. Actually, the precise point where you made an error is by putting the limit on the outside again - once you turned it into the indeterminate form, your statement is equivalent to the fact that the limit is equal to the ratio of two other limits. That part is technically true. Unfortunately, they both turn out to be 0 - so you are effectively back to 0/0 - the information is lost, so it is actually a dead end.

4. ## Re: Finding the limit of a particular trig function as x approaches 0

I did approach it in the conventional way as well as L'Hopital's rule and got -1. I was curious as to why substitution failed, and whether it had been an error in my part.

5. ## Re: Finding the limit of a particular trig function as x approaches 0

Well, the word "effectively" in your first post is where the logic breaks down. Once the limited is evaluated, its a number, you can't transform a particular number back into a limit and move the limit sign that way...

6. ## Re: Finding the limit of a particular trig function as x approaches 0

Originally Posted by SworD
Well, the word "effectively" in your first post is where the logic breaks down. Once the limited is evaluated, its a number, you can't transform a particular number back into a limit and move the limit sign that way...
Yes, I understood after your first post. When I said "effectively", I don't mean that it follows from the previous step, rather that it can be seen that it's evaluation would be equivalent to $\lim_{x\to\0}\frac{sinx}{x}=1$; when you substitute 0 into x, it would yield the same "number" as when you substitute 0 into x for the original problem. However it's already been pointed out by you that x and u do not approach 0 at the same place.

Thanks again to you both.