#lim_(t->oo)(sqrt(t)-1)/(t-1) = # ?

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Answer 1 · 2016-09-20T00:46:47

I think it was for #t->oo#...giving zero in the limit.

If it is #t->oo#:

#lim_(t->oo)(sqrt(t)-1)/(t-1)=oo/oo#

To solve this we can use de L'Hospital Rule deriving top and bottom to get:
#lim_(t->oo)(1/(2sqrt(t)))/1=lim_(t->oo)1/(2sqrt(t))#

as #t->oo# the argument tends to zero;

#lim_(t->oo)1/(2sqrt(t))=0#

Answer 2 · 2016-09-23T07:45:32

#lim_(x->oo)(sqrt(t)-1)/(t-1)=0#

An alternative method, solving algebraically:

#lim_(x->oo)(sqrt(t)-1)/(t-1)#

#=lim_(x->oo)((sqrt(t)-1)(sqrt(t)+1))/((t-1)(sqrt(t)+1))#

#=lim_(x->oo)(t-1)/((t-1)(sqrt(t)+1))#

#=lim_(x->oo)1/(sqrt(t)+1)#

#=1/oo#

#=0#

Answer 3 · 2016-09-23T07:55:34

Making #y = sqrt t#

#lim_(t->oo)(sqrt(t)-1)/(t-1) equiv lim_(y->oo)(y-1)/(y^2-1) = lim_(y->oo)1/(y+1) = 0#