How do you find the limit #sin(2x)/ln(x+1)# as #x->0#?

Question

7561 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-01T08:55:42

#lim_(x->0) sin(2x)/ln(1+x) =2#

As the limit is in the indeterminate form #0/0# we can use l'Hospital's rule saying that in such case:

#lim_(x->0) sin(2x)/ln(1+x) = lim_(x->0) (d/dxsin(2x))/(d/dxln(1+x))#

#lim_(x->0) sin(2x)/ln(1+x) = lim_(x->0) (2cos(2x))/(1/(1+x)) =2#

Alternatively we can remember two important limits:

#lim_(x->0) sinx/x = 1#

#lim_(x->0) ln(1+x)/x = 1#

so that if we write the function as:

#lim_(x->0) sin(2x) / ln(1+x) = lim_(x->0) sin(2x)/(2x) * 2 * x/ln(1+x) =2#