How do I evaluate without using L'Hopital's rule lim sqrt(x^2-9)/abs(3-x) as x->3- I didn't manage to manipulate the algebric expression. I got to sqrt((x+3)/(x-3)) as x->3- and I don't know what to do from here
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Originally Posted by GIPC How do I evaluate without using L'Hopital's rule lim sqrt(x^2-9)/abs(3-x) as x->3- I didn't manage to manipulate the algebric expression. I got to sqrt((x+3)/(x-3)) as x->3- and I don't know what to do from here Since you are approaching from the left, you need to take notice of what happens with the values of . Notice that if . So . This now tends to which tends to .
and how did you conclude that 6/0 tends to infinity? I can't use L'Hopital's rule in this one, how can I explain this result otherwise?
Originally Posted by GIPC and how did you conclude that 6/0 tends to infinity? I can't use L'Hopital's rule in this one, how can I explain this result otherwise? How many 0's go into 6? 0 + 0 + 0 + ... = 0 So infinitely many will go into 6.
I understand the intuitive explanation but if I had to submit the homework, How do I show it on a more formal note?
Originally Posted by GIPC I understand the intuitive explanation but if I had to submit the homework, How do I show it on a more formal note? It's well known that for . Check the graph of . It's one of the most basic limits and one you will need to remember.
Last edited by Prove It; April 14th 2010 at 03:55 AM.
Originally Posted by Prove It It's well known that for . Check the graph of . It's one of the most basic limits and one you will need to remember. Typo- that should be .
Originally Posted by HallsofIvy Typo- that should be . Thanks, will edit now.
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