I finding the limits of trig functions

[Oomair]

Homework Statement

lim cos(beta sign)-1/sin(beta)
Beta-0

2. lim sin^2 3t/t^2
t-0

for the first one i tried to use the quotient formula to get the derivative, but still I am not sure i did it correctly and for the second problem, i have no idea what to do

Homework Equations

The Attempt at a Solution

 

[Coolphreak]

in a way this might be cheating (depending on how your teacher wants you to solve the problem), but have u tried l'hopital's rule?

 

[mjsd]

for the second one you can try this: expand Sin^2 (3t) as a power series... then you will find that the first term is proportional to t^2 and after that the answer is obvious

 

[VietDao29]

Homework Statement

lim cos(beta sign)-1/sin(beta)
Beta-0
...
for the first one i tried to use the quotient formula to get the derivative, but still I am not sure i did it correctly

What do you mean by Quotient Formula? =.="
Quotient Formula is used for taking the derivative of an expression, but, in this problem, you are asked to find the limit of an expression.

Can you show us what you did?

2. lim sin^2 3t/t^2
t-0

Do you know this limit:
$$\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$?

When seeing an expression with sin(x) over x, or something like that, you should think about using this well-known limit right away. So, you can re-arrange the expression a bit, like this:

$$\lim_{t \rightarrow 0} \frac{\sin ^ 2 (3t)}{t ^ 2} = \lim_{t \rightarrow 0} \frac{9 \sin ^ 2 (3t)}{(3t) ^ 2} = ...$$

Can you go from here? :)

 

[Oomair]

^yep that's what i was looking for, but why did u multiply a 9 instead of a 3?

 

[malawi_glenn]

Because in the denominator, he also multiplied with 9, but wrote it as 9 = 3^2, the denominator must be exactly the same as the argument of the sinus.