How do you find #lim ln(t^2+1)/t# as #t->0# using l'Hospital's Rule?
1 Answer
Explanation:
L'Hospital's Rule can be used when we have a limit of and indeterminate form of
So if
# lim_(x rarr a)f(x)/g(x) = lim_(x rarr a)(f′(x))/(g′(x)) #
For
As
So:
#lim_(t rarr 0)ln(t^2+1)/t = lim_(t rarr 0) (d/dt(ln(t^2+1)))/(d/dt(t)) #
# " " = lim_(t rarr 0) (1/(t^2+1)*2t)/1#
# " " = lim_(t rarr 0) (2t)/(t^2+1) #
# " " = 0 #
Which we can confirm graphically:
graph{ln(x^2+1)/x [-10, 10, -5, 5]}