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 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