# How do you find #lim (t+1/t)((4-t)^(3/2)-8)# as #t->0# using l'Hospital's Rule?

##### 1 Answer

Aug 14, 2017

Please see below.

#### Explanation:

We will rewrite it as

Note that

# = lim_(trarr0)((4-t)^(3/2)-8)/(t/(t^2+1))#

# = lim_(trarr0)(-(3/2)(4-t)^(1/2))/((1(t^2+1)-t(2t))/(t^2+1)^2)#

# = lim_(trarr0)(-(3/2)(4-t)^(1/2))/((1-t^2)/(t^2+1)^2)#

# = ((-3/2)(2))/(1/1^2) = -3#