# 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]}