Stuck on a possibly simple trig limit

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Homework Statement



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 I am stuck ? i may be way over complicating it but i think it needs some algebraic manipulation in the current form?

Homework Equations





The Attempt at a Solution

 
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You should notice that sin \textit{x} /\textit{x} is a notable limit
 
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.
 
[STRIKE][STRIKE][/STRIKE][/STRIKE]yes I'm aware of sinx/x = 1 and all the other variations of it etc that's what i was referring 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 I am just doing something stupid with my algebra any pointers on that one?

rember its 2t / (sin(t)) - t
 
2t/ sin(t) - t
 
If you mean

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

then rewrite it in the form

\frac{2}{\frac{\sin t}{t} -1}.
 
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)
 
that was exactly what i was doing wrong.. thank you very much

it becomes

forgive my poor latex.. I am trying

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

that is 2 / 1 - t/t and since t/t is approaching zero but both equal shouldn't 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 don't see the infinity situation here?
 
rambo5330 said:
that was exactly what i was doing wrong.. thank you very much

it becomes

forgive my poor latex.. I am trying

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

that is 2 / 1 - t/t and since t/t is approaching zero but both equal shouldn't 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 don't see the infinity situation here?

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
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
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
excellent thanks for your help!... i did look in the textbook 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!
 
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