# How do you find lim (1/t-1)/(t^2-2t+1) as t->1^+ using l'Hospital's Rule?

##### 1 Answer
Jul 21, 2017

Te limit does not exist

#### Explanation:

We seek:

$L = {\lim}_{t \rightarrow {1}^{+}} \frac{\frac{1}{t} - 1}{{t}^{2} - 2 t + 1}$

From the graph it would appear that the limit does not exist:

graph{(1/x-1)/(x^2-2x+1) [-5, 5, -6, 6]}

As this is of an indeterminate form $\frac{0}{0}$ we can apply L'Hôpital's rule.

$L = {\lim}_{t \rightarrow {1}^{+}} \frac{\frac{d}{\mathrm{dt}} \left(\frac{1}{t} - 1\right)}{\frac{d}{\mathrm{dt}} \left({t}^{2} - 2 t + 1\right)}$

$\setminus \setminus = {\lim}_{t \rightarrow {1}^{+}} \frac{- \frac{1}{t} ^ 2}{2 t - 2}$

$\setminus \setminus \rightarrow \infty$, as predicted