Find the limit?

Question

Find lim t→∞ (4t²(sin(2/t))²).

 

Answer: 16

Answer 1

The limit is of the indeterminate form 0·∞. To use l'Hospital's Rule, bring it into the standard indeterminate form 0/0:

lim _t→∞ (4t²(sin(2/t))²) =4 lim _t→∞ ((sin(2/t))²/(1/t²))

 

You need to use l'Hospital's Rule twice now:

 

4 lim _t→∞ ((sin(2/t))²/(1/t²)) = 4 lim (2*sin(2/t)*cos(2/t)*(-2/t²)/(-2/t³)) (1st time)

=4 lim (sin(4/t)/(1/t))

=4 lim (cos(4/t)*(-4/t²)/(-1/t²))  (2nd time)

=16 lim _t→∞ (cos(4/t))

=16

Answer 2

lim _t→∞ (4t²(sin²(2/t)))

let u=2/t and so t=2/u and t²=4/u² as t→∞ so u→0

 

lim_u→0(16/u²)sin²u=16lim_u→0(sin²u/u²)=16×1=16 by the known value of lim_x→0(sinx/x)=1