# Stuck on a possibly simple trig limit

1. Sep 29, 2010

### rambo5330

1. The problem statement, all variables and given/known data

how do you go about solving the equation
lim (t->0) 2t / (sin(t)) - t

the answer in the text is significantly different than what i get.. i can get most of the other trig limits using the fundamental limit etc.. but this one im stuck ? i may be way over complicating it but i think it needs some algebraic manipulation in the current form?

2. Relevant equations

3. The attempt at a solution

2. Sep 29, 2010

### Quinzio

You should notice that $$sin \textit{x} /\textit{x}$$ is a notable limit

3. Sep 29, 2010

### Lanthanum

lim x->0 of sinx/x =1, so
lim x->0 of 1/(sinx/x) is also 1, and 1/(sinx/x)=x/sinx, so
lim x->0 of x/sinx = 1
from here just rearrange the equation.

4. Sep 30, 2010

### rambo5330

[STRIKE][STRIKE][/STRIKE][/STRIKE]yes I'm aware of sinx/x = 1 and all the other variations of it etc thats what i was refering to as the fundamental trig limit but i think its the fact its in the form. sinx - x i cannot factor out an x and since its in the form of a binomial this is where i think im just doing something stupid with my algebra any pointers on that one?

rember its 2t / (sin(t)) - t

5. Sep 30, 2010

### rambo5330

$$2t/ sin(t) - t$$

6. Sep 30, 2010

### fzero

If you mean

$$\frac{2t}{\sin t -t},$$

then rewrite it in the form

$$\frac{2}{\frac{\sin t}{t} -1}.$$

7. Sep 30, 2010

### Lanthanum

If you factor out the 2 and split the limit into two parts then you have;
2 lim t->0 (t/sint) - lim t->0 (t)

8. Sep 30, 2010

### rambo5330

that was exactly what i was doing wrong.. thank you very much

it becomes

forgive my poor latex.. im trying

$$2 / sint/t - t/t$$
sorry ill have to figure out better latex tomorrow
but if im correect

that is 2 / 1 - t/t and since t/t is approaching zero but both equal shouldnt that become 1 as well... giving 1-1 in the denominator and an undefined situation?? the text is saying it should equal DNE or +infinity .... i dont see the infinity situation here?

9. Sep 30, 2010

### fzero

Well

$$\frac{1}{0}=\infty$$

in the sense that

$$\lim_{x\rightarrow 0} \frac{1}{x} \rightarrow \infty.$$

So the limit does not exist.

10. Sep 30, 2010

### rambo5330

but in that form it just purely does not exist am i right? unless you know wether you're approaching zero from the right or the left.. you can end up with -infinity or +infinity....

so it would need to be a one sided limit.. or am i off?

11. Sep 30, 2010

### fzero

You're right if the left and right limits don't agree, the limit does not exist. However, it's also true that a divergent result means that the limit doesn't exist.

In this case, the function is even, so both the right and left limits give $$-\infty$$. We still say that the limit does not exist.

12. Sep 30, 2010

### rambo5330

excellent thanks for your help!... i did look in the text book again to confirm the answer and i see its showing a +infinity?? not sure what the reasoning behind the positive is? but fighting with this question has actually helped me understand a bit deeper. thanks!