Question

find the limits using L'Hospital's rule is possible:

lim (sin pix)/(x^2-16)
x->4

Answer 1

Answer:

1/2

Step-by-step explanation:

$\lim_{x \to \ 4}[\frac{sin \pi x }{x^2-16} ]$

if we put x=4 we get an undetermined limit that is 0/0 so we differentiate the numerator and denominator separately

$\lim_{x \to \ 4}[\frac{\pi cos \pi x }{2x} ]$

then we insert the limit x=4 we get

$\lim_{x \to \ 4}[\frac{\pi cos 4\pi }{8} ]$

so we get the

$4[\frac{1}{8}]$

final answer is $\frac{1}{2}$

so that would be the limit.