#lim_(x to 0) (ln(1 + 3x) )/ sin(x)#

For starters, we can see that if we set #x = 0# we are looking at:

#ln(1)/sin(0) = 0/0#.

That is in indeterminate form and so L'Hôpital's Rule can be used. That can be a bit of a mechanical black-box and results, after one application, in this:

#= lim_(x to 0) (3/(1 + 3x) )/ cos(x) = 3#

If you are familiar with Taylor Series, we can look at this in a bit more detail:

#\sin (x) \approx x- x^{3}/{3!}+x^{5}/{5!}- x^{7}/{7!}#

....and...

#log(1+x)=x- x^{2}/2+ x^{3}/{3}+ O(x^{4})#

Or:

#log(1+3x)=3x- (3x)^{2}/2+ (3x)^{3}/{3}+ O(x^{4})#

As #x to 0#, we can ignore #x^2# higher order terms and compare the first terms (in #x^1#) to conclude the same thing, namely that:

#lim_(x to 0) (ln(1 + 3x) )/ sin(x) = lim_(x to 0) (3x) / x = 3#

It might then be helpful to show how the Taylor Series connect into L'Hôpital, if we compare series for 2 functions, f(x) and g(x), about points at which #f(a) = g(a) = 0#.

They create the following ratio:

#(f(x))/(g(x)) = (f(a)+ {f'(a)}/{1!}(x-a)+ {f''(a)}/{2!}(x-a)^{2}+\cdots )/(g(a)+ {g'(a)}/{1!}(x-a)+ {g''(a)}/{2!}(x-a)^{2}+\cdots)#

For #f(a) = g(a) = 0# and #x approx a# you can see already that the predominant terms in this ratio are the first derivatives.