# Question #18c98

Mar 6, 2017

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

${\lim}_{x \to 0} \frac{\left(\sin \left(5 x\right)\right) \left(1 - \cos \left(7 x\right)\right)}{\left({2}^{x} - {3}^{x}\right) \left(\log \left(1 + 4 x\right)\right)}$
By direct substitution
$= \frac{0}{0}$

${\lim}_{x \to 0} \frac{\left(5 \cos \left(5 x\right)\right) \left(1 - \cos \left(7 x\right)\right) + \left(\sin \left(5 x\right)\right) \left(7 \sin \left(7 x\right)\right)}{\left({2}^{x} \ln 2 - {3}^{x} \ln 3\right) \left(\log \left(1 + 4 x\right)\right) + \frac{4 \left({2}^{x} - {3}^{x}\right)}{\left(1 + 4 x\right) \left(\ln 10\right)}}$

${\lim}_{x \to 0} \frac{5 \cos \left(5 x\right) - 5 \cos \left(5 x\right) \cos \left(7 x\right) + 7 \sin \left(5 x\right) \sin \left(7 x\right)}{{2}^{x} \left(\ln 2\right) \left(\log \left(1 + 4 x\right)\right) - {3}^{x} \left(\ln 3\right) \left(\log \left(1 + 4 x\right)\right) + \frac{4 \left({2}^{x} - {3}^{x}\right)}{\left(1 + 4 x\right) \left(\ln 10\right)}}$

$= \frac{0}{0}$

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

And the limit does approach 0.