# 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

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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 #