lim x -> 0 (cos(5x) - 1)/sin(7x)

Question

lim x -> 0 (cos(5x) - 1)/sin(7x)

Answer 1

If you have not learned  L'Hopital's Rule yet, we can make use of the following limits commonly discussed in Calculus classes:

 

   lim_x→0(sinx/x) = 1 and lim_x→0[(cosx-1)/x] = 0

 

So, lim_x→0[(cos(5x)-1)/sin(7x)]

 

   = lim_x→0[{(cos(5x)-1)/x}/{sin(7x)/x}]

 

   = lim_x→0[{5(cos(5x)-1)/(5x)}/{7sin(7x)/(7x)}]

 

         As x→0, 5x→0 and 7x→0

 

So, evaluation of the limit above yields:  [(5)(0)]/[(7)(1)] = 0

 

Answer 2

Limit as x→0 of (cos5x -1) / sin7x = ?

 

Use L'Hopital's rule:

 

Limit as x→0 of (cos5x -1) / sin7x = Limit as x→0 of (-5sin5x) / (7cos7x) = 0

 

The limit is 0.