Evaluate the lim x-0 for (1-cos5x)/sin5x
I can't use l'hopitals rule but i am told to use these theorems "as needed" :
lim x-0 for sinx/x = 1 and lim x-0 for (1-cosx)/x =0

Question

Evaluate the lim x->0 for (1-cos5x)/sin5x 
  I can't use l'hopitals rule but i am told to use these theorems "as needed" :
lim x->0 for sinx/x = 1  and lim x->0  for (1-cosx)/x =0

anonymous · Answer

Thanks for answering again! I have gotta run to class so I can't answer any questions about it, but would still love a reply! Thanks!

zehanz · Answer

Hello again!
I think you'll need both of these standard limits: The limit you've got to evaluate is a mixture of both. 
Hint:$$\frac{ a }{ b }=\frac{ a }{ c }\cdot \frac{ c }{ b }$$Or am I too cryptic here?

zehanz · Answer

So you can write the fraction as$$\frac{ 1-\cos5x }{ 5x }\cdot \frac{ 5x }{ \sin5x }$$And then calculate the product of two separate limits. You will still need to handle the "5x" instead of x, and the second part being upside down, but I think you'll be able to do that!