# 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 ${\lim}_{t \to 0} \left(1 - {\left(1 + t\right)}^{2} / \left(t {\left(1 + t\right)}^{2}\right)\right)$

May 28, 2018

$\textcolor{b l u e}{{\lim}_{t \to 0} \left(\frac{1 - {\left(1 + t\right)}^{2}}{t {\left(1 + t\right)}^{2}}\right) = {\lim}_{t \to 0} \frac{- 2 \left(1 + t\right)}{t \cdot 2 \left(1 + t\right) + {\left(1 + t\right)}^{2}} = - \frac{2}{1} = - 2}$

#### Explanation:

I assume it is ${\lim}_{t \to 0} \left(\frac{1 - {\left(1 + t\right)}^{2}}{t {\left(1 + t\right)}^{2}}\right)$

${\lim}_{t \to 0} \left(\frac{1 - {\left(1 + t\right)}^{2}}{t {\left(1 + t\right)}^{2}}\right) = \frac{0}{0}$

since the direct compensation product equal $\frac{0}{0}$
we will use L'hospital Rule.
L'hospital Rule $\textcolor{red}{{\lim}_{t \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}}$

now lets applied L'hospital Rule.

${\lim}_{t \to 0} \left(\frac{1 - {\left(1 + t\right)}^{2}}{t {\left(1 + t\right)}^{2}}\right) = {\lim}_{t \to 0} \frac{- 2 \left(1 + t\right)}{t \cdot 2 \left(1 + t\right) + {\left(1 + t\right)}^{2}} = - \frac{2}{1} = - 2$

May 28, 2018

${\lim}_{t \to 0} \frac{1 - {\left(1 + t\right)}^{2}}{t {\left(1 + t\right)}^{2}} = - 2$

#### Explanation:

If we seek the value of the corrected limit:

$L = {\lim}_{t \to 0} \frac{1 - {\left(1 + t\right)}^{2}}{t {\left(1 + t\right)}^{2}}$

Then by expanding the numerator, we have:

$L = {\lim}_{t \to 0} \frac{1 - \left(1 + 2 t + {t}^{2}\right)}{t {\left(1 + t\right)}^{2}}$

$\setminus \setminus = {\lim}_{t \to 0} \frac{- 2 t - {t}^{2}}{t {\left(1 + t\right)}^{2}}$

$\setminus \setminus = {\lim}_{t \to 0} \frac{- t \left(2 + t\right)}{t {\left(1 + t\right)}^{2}}$

$\setminus \setminus = {\lim}_{t \to 0} \frac{- \left(2 + t\right)}{{\left(1 + t\right)}^{2}}$

And we can evaluate this limit by direct substitution:

$L = \frac{- \left(2 + 0\right)}{{\left(1 + 0\right)}^{2}}$
$\setminus \setminus = - 2$