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

Question

9660 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-08T02:14:41

Use L'Hôpital's rule and evaluate the resulting expression at 0.

Given: #lim_(xto0)sin(x)/(x + sin(x)) = ?#

Because the expression evaluated at 0, is the indeterminate form, #0/0#, the use of L'Hôpital's rule is warranted.

Compute the derivative of the numerator:

#(d(sin(x)))/dx = cos(x)#

Compute the derivative of the denominator:

#(d(x + sin(x)))/dx = 1 + cos(x)#

Take the limit of the new fraction:

#lim_(xto0)cos(x)/(1 + cos(x)) = cos(0)/(1 + cos(0)) = 1/2#

According to L'Hôpital's rule, the limit of the original function goes to the same value:

#lim_(xto0)sin(x)/(x + sin(x)) = 1/2#