Ask your own question, for FREE!
Calculus1 23 Online
OpenStudy (anonymous):

I need help with the steps involved to solve the following equation. lim x→−24 (square root(x^2 + 49) − 25)/x + 24 obviously when i plug it in its 0/0 but I'm unsure of the method to solve at this point. if there wasnt a square root i could try factoring and reducing, but with the root in there i don't know how to proceed. thanks!

Parth (parthkohli):

Do you know l'hopital's rule?

Parth (parthkohli):

That should really be enough, in my opinion.

Parth (parthkohli):

Or the definition of derivative.

Parth (parthkohli):

Assume \(h =x + 24 \), then the limit is written as the following:\[\lim_{h\to 0} \left(\sqrt{x^2 + 49} - 25 \over h\right) \]

Parth (parthkohli):

Hmm, the definition of derivative doesn't apply here. NVM

OpenStudy (anonymous):

surprisingly enough, we havent covered derivatives yet. or that rule

Parth (parthkohli):

Darn.

OpenStudy (anonymous):

I haven't had math in years and I'm pretty lost in this class. what we have covered are the limit laws; sum, difference, constant multiple, product, quotient

OpenStudy (shubhamsrg):

You should rationalize the numerator. Must help.

Parth (parthkohli):

Yay, rationalizing helped me. Shubham, Y U NO REPLY TWO SECONDS LATER?

OpenStudy (shubhamsrg):

I may delete my comment if you ask for it! ;)

Parth (parthkohli):

No, you have your credit.

Parth (parthkohli):

:-P

OpenStudy (anonymous):

try \[\frac{\sqrt{x^2+49}-25}{x-24}\times \frac{\sqrt{x^2-49}+25}{\sqrt{x^2-49}+25}\]

Parth (parthkohli):

Yes, that's what I did there. Conjugation for the win.

OpenStudy (anonymous):

ok that was wrong!!

Parth (parthkohli):

No it wasn't?

OpenStudy (anonymous):

\[\frac{\sqrt{x^2+49}-25}{x+24}\times \frac{\sqrt{x^2+49}+25}{\sqrt{x^2+49}+25}\] is more like it

OpenStudy (anonymous):

I tried that, which gets me \[\frac{ (x^2+49)-625 }{ (x+24) * (\sqrt{x^2+49} +25) }\] Or I think it does. And I'm not sure where to go from there.

OpenStudy (shubhamsrg):

-49+625 = -576 = -(24^2) Does it help ?

OpenStudy (anonymous):

yes! I'm so bad at factoring, I dont ever seem to see stuff like that

OpenStudy (shubhamsrg):

Glad to have helped! ^_^

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!