Question #18c98

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Answer 1 · 2017-03-06T05:34:05

I'm not sure if I interpreted your placement of parentheses correctly, so I apologize if the original expression is incorrect.

#lim_(x->0)frac{(sin(5x))(1-cos(7x))}{(2^x-3^x)(log(1+4x))}#
By direct substitution
#=0/0#

#lim_(x->0)frac{(5cos(5x))(1-cos(7x))+(sin(5x))(7sin(7x))}{(2^xln2-3^xln3)(log(1+4x))+frac{4(2^x-3^x)}{(1+4x)(ln10)}}#

#lim_(x->0)frac{5cos(5x)-5cos(5x)cos(7x)+7sin(5x)sin(7x)}{2^x(ln2)(log(1+4x))-3^x(ln3)(log(1+4x))+frac{4(2^x-3^x)}{(1+4x)(ln10)}}#

#=0/0#

At this point we could do L'Hospitals rule again (multiple more times probably), but I just decided to graph it on Desmos

enter image source here

And the limit does approach 0.