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

1 Answer
Jim H
Aug 14, 2017

Please see below.

Explanation:

#(t+1/t)((4-t)^(3/2)-8)# as #trarr0# has form #oo*0#.

We will rewrite it as #((4-t)^(3/2)-8)/(1/(t+1/t)# so we have an expression whose limit has form #0/0# and we can use l'Hospital's Rule.

Note that #t+1/t = (t^2+1)/t# so we want

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

# = 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#

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