How do you find the limit sin(2x)/ln(x+1)sin(2x)ln(x+1) as x->0x0?

1 Answer
Apr 1, 2017

lim_(x->0) sin(2x)/ln(1+x) =2limx0sin(2x)ln(1+x)=2

Explanation:

As the limit is in the indeterminate form 0/000 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))limx0sin(2x)ln(1+x)=limx0ddxsin(2x)ddxln(1+x)

lim_(x->0) sin(2x)/ln(1+x) = lim_(x->0) (2cos(2x))/(1/(1+x)) =2limx0sin(2x)ln(1+x)=limx02cos(2x)11+x=2

Alternatively we can remember two important limits:

lim_(x->0) sinx/x = 1limx0sinxx=1

lim_(x->0) ln(1+x)/x = 1limx0ln(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