Find the limit by evaluating the derivative of a suitable function at an appropriate point? lim t approaches 0 (1-(1+t)^2/t(1+t)^2) Thank you very much

Do you mean #lim_(t to 0)(1-(1+t)^2/(t(1+t)^2))#

2 Answers
May 28, 2018

#color(blue)[lim_(t to 0)((1-(1+t)^2)/(t(1+t)^2))=lim_(t to 0)[-2(1+t)]/[t*2(1+t)+(1+t)^2]=-2/1=-2]#

Explanation:

I assume it is #lim_(t to 0)((1-(1+t)^2)/(t(1+t)^2))#

#lim_(t to 0)((1-(1+t)^2)/(t(1+t)^2))=0/0#

since the direct compensation product equal #0/0#
we will use L'hospital Rule.
L'hospital Rule #color(red)[lim_(trarra)(f'(x))/(g'(x))]#

now lets applied L'hospital Rule.

#lim_(t to 0)((1-(1+t)^2)/(t(1+t)^2))=lim_(t to 0)[-2(1+t)]/[t*2(1+t)+(1+t)^2]=-2/1=-2#

May 28, 2018

# lim_(t to 0) (1-(1+t)^2)/(t(1+t)^2) = -2#

Explanation:

If we seek the value of the corrected limit:

# L = lim_(t to 0) (1-(1+t)^2)/(t(1+t)^2) #

Then by expanding the numerator, we have:

# L = lim_(t to 0) (1-(1+2t+t^2))/(t(1+t)^2) #

# \ \ = lim_(t to 0) (-2t-t^2)/(t(1+t)^2) #

# \ \ = lim_(t to 0) (-t(2+t))/(t(1+t)^2) #

# \ \ = lim_(t to 0) (-(2+t))/((1+t)^2) #

And we can evaluate this limit by direct substitution:

# L = (-(2+0))/((1+0)^2) #
# \ \ = -2 #