problem with limity

Question

+5

1041

1

+9

hi everyone :)

how could I solve this :

WITHOUT USING L'HOSPITAL RULE :)

IM LOST :(

skcmarek Oct 27, 2014

+33615

+5

Best Answer

When x is very large pi/5^x is very small.

When z is very small sin(z) → z, so when x is very large xsin(pi/5^x) → x*pi/5^x

5^x increases much more rapidly than x, so x*pi/5^x → 0

Hence, in the limit as x → ∞, xsin(pi/5^x) → 0

.

Alan Oct 27, 2014

Alan · Answer 1 · 2014-10-27T09:18:58

When x is very large pi/5^x is very small.

When z is very small sin(z) → z, so when x is very large xsin(pi/5^x) → x*pi/5^x

5^x increases much more rapidly than x, so x*pi/5^x → 0

Hence, in the limit as x → ∞, xsin(pi/5^x) → 0

.