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
# 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
