What is the #lim_{t->0} 1/(t sqrt(1+t)) - 1/t#?

1 Answer
Sep 24, 2016

-1/2

Explanation:

STEP 1: Determine the common denominator
The common denominator is #t sqrt(1+t)#

STEP 2: Rewrite with the common denominator
#lim_{t->0} 1/(t sqrt(1+t)) - sqrt(1+t)/(t sqrt(1+t)#
#=lim_{t->0} (1-sqrt(1+t))/(t sqrt(1+t)#

STEP 3: Now, we feel a little stuck. So let's try multiplying by the conjugate to get rid of the square root in the numerator.
#lim_{t->0} (1-sqrt(1+t))/(t sqrt(1+t)) * (1+sqrt(1+t))/(1+sqrt(1+t))#
=#=lim_{t->0} (1-(1+t))/(t sqrt(1+t)(1+sqrt(1+t))#
=#=lim_{t->0} -t/(t sqrt(1+t)(1+sqrt(1+t))#

STEP 4: Cancel the t in the numerator and denominator:
=#=lim_{t->0} -1/(sqrt(1+t)(1+sqrt(1+t))#

STEP 5: Now that we no longer have a 0 in the denominator, we can evaluate by plugging the 0 in for t:
#=-1/(sqrt(1+0)(1+sqrt(1+0))) = -1/(1+1) = -1/2#