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))#
Do you mean
2 Answers
Explanation:
I assume it is
since the direct compensation product equal
we will use L'hospital Rule.
L'hospital Rule
now lets applied L'hospital Rule.
# 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 #