you do have a 0/0 form
Using Hospital's rule
then lim= (e^x+e^-x-2)/(x+cosx)
Lim= 0/1=0
Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim x→0 (e^x − e^−x − 2x)/(x − sin(x))
2 answers
using L'Hospital's rule we have
lim x→0 (e^x − e^−x − 2x)/(x − sin(x))
lim x→0 (e^x + e^−x − 2)/(1 − cosx)
still 0/0, so do it again:
lim x→0 (e^x - e^−x)/sinx
and again:
lim x→0 (e^x + e^−x)/cosx
→ (1+1)/1 = 2
lim x→0 (e^x − e^−x − 2x)/(x − sin(x))
lim x→0 (e^x + e^−x − 2)/(1 − cosx)
still 0/0, so do it again:
lim x→0 (e^x - e^−x)/sinx
and again:
lim x→0 (e^x + e^−x)/cosx
→ (1+1)/1 = 2