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

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Answer 1 · 2017-07-21T13:07:42

Te limit does not exist

We seek:

$L = lim_(t rarr 1^+) (1/t-1)/(t^2-2t+1)$

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 $0/0$ we can apply L'Hôpital's rule.

$L = lim_(t rarr 1^+) (d/dt(1/t-1) )/( d/dt (t^2-2t+1) )$

$\ \ = lim_(t rarr 1^+) (-1/t^2 )/( 2t-2 )$

$\ \ rarr oo$ , as predicted

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