Question

I'm stumped on how to approach lim( sqrt{t+1} - sqrt{t-1} )/t, as t approaches 0. Can you get me started?

Answer 1

\[\lim_{t \rightarrow 0} (\sqrt{t+1}-\sqrt{t-1})/t\]

Answer 2

it is 0/0 form..so you can use l'hospital rule..differnetiate the numerator and denominator ...to reduce it

Answer 3

\[=1/(2\sqrt {t+1} )-1/(2\sqrt{t-1})\]

Answer 4

can u solve nw

Answer 5

We have not yet covered l'hopital's Rule or Derivatives yet. That is the next section. Is there another way to simplify it that i'm just not seeing?

Answer 6

this is not 0/0 form. The numerator takes a complex form and denominator tends to 0. So, it tends to complex infinity when t tends from 0+ and -complex infinity when t tends from 0-. I do not know what complex infinity means though. However, it is safe to say that limit doesn't exist.

Answer 7

Thank you for your help.

Answer 8

Hmmm... One idea that I had (appreciate any corrections!) is to do a bit of algebra before applying the limit operation. Consider taking apart the expression \[(\sqrt{t+1} - \sqrt{t-1})/t\] as \[\sqrt{t+1}/t - \sqrt{t-1}/t\] . Then, for the first term, multiply numerator and denominator by \[\sqrt{t+1}\] This then gives you: \[(t+1)/t \sqrt{t+1}\] . Now, multiply both numerator and denominator by 1/t and that gives you for the first term: \[(1+1/t)/\sqrt{t+1}\] Then, we work on the second term in a similar fashion, first multiplying numerator and denominator by \[\sqrt{t-1}\] and then multiply the numerator and denominator by 1/t - then you end up with a expression that can be more readily evaluated with the limit operation: \[(1+1/t)/\sqrt{t+1} - (1-1/t)/\sqrt{t-1}\]That should be equal to 0 when t approaches 0. Does that look OK? You don't need L'Hospitals rule, it looks like algebra can do the job.

Answer 9

@gwicks well..... how does that equal 0?

Answer 10

@nipunmalhotra93: You're right! Bad algebra on my part - the 1/t terms would in fact go towards infinity as t->0. Good catch, thank you. I suspect though, that there is an algebraic way. I'll have to look at it some more. thanks again.

Answer 11

Ahh, the obvious escaped me entirely: the term \[\sqrt{t+1}\]is of course continuous at zero, HOWEVER, the term \[\sqrt{t-1}\] has no Real values for any t < +1 since the term becomes complex, so therefore the expression is not real valued at t=0. So, @nipunmalhotra93 was very much on target.