How do you find the limit of sinx/(5x) as x->0?

2 Answers
Oct 25, 2016

The limit is 1/5

Explanation:

Straight forward sinx/(5x)=0 when x->0

So we have to apply l'Hôpital's rule

The limit is ((sinx)')/((5x)')=cosx/5

And limit as x->0 of cosx/5=cos0/5=1/5

Oct 25, 2016

Rewrite it so you can use the fundamental trigonometric limit lim_(thetararr0)sintheta/theta = 1

Explanation:

sinx/(5x) = 1/5(sinx/x)
So

lim_(xrarr0)sinx/(5x) = lim_(xrarr0)1/5(sinx/x)

= 1/5 lim_(xrarr0)sinx/x

= 1/5(1) = 1/5