thinay Posts: 40, Reputation: 2 Junior Member #1 Jul 8, 2009, 09:03 AM
Can anyone help me in my problem again? Here's my problem in limits:

Lim ((Square root of 7 + a) - square root of 7) over a
a->0

Hope you can help me. Thanks in advance. :)
 galactus Posts: 2,271, Reputation: 282 Ultra Member #2 Jul 8, 2009, 09:14 AM
Finally. A problem besides "solve x+1=0 for x".:)

$\lim_{a\to 0}\frac{\sqrt{7+a}-\sqrt{7}}{a}$

Multiply the top and bottom by the conjugate of the numerator.

$\frac{(\sqrt{7+a}-\sqrt{7})}{a}$$\cdot\frac{(\sqrt{7+a}+\sqrt{7})}{(\sqrt{7+a}+ \sqrt{7})}$

This gives:

$\lim_{a\to 0}\frac{\not{a}}{\not{a}(\sqrt{7+a}+\sqrt{7})}$

Now, can you see what you get as a-->0? Tell me what you get.
 thinay Posts: 40, Reputation: 2 Junior Member #3 Jul 8, 2009, 04:32 PM

I end up with an answer:
1/square root of 14

Is my answer correct? Because the hand out that was given to us tells that the answer is (square root of 7)/14.

How did they get that answer? There is no solution given in my hand out, that's why I'm trying to solve this one on my own and with your help. :)
 galactus Posts: 2,271, Reputation: 282 Ultra Member #4 Jul 9, 2009, 05:12 AM

How did you get that? Just bu plugging in a=0 in the limit, we get $\frac{1}{\sqrt{0+7}+\sqrt{7}}=\frac{1}{2\sqrt{7}}$

The solution should be $\frac{1}{2\sqrt{7}}=\frac{\sqrt{7}}{14}$
 thinay Posts: 40, Reputation: 2 Junior Member #5 Jul 9, 2009, 08:12 AM

Oh.. Stupid me. :D now I know my mistake. When I add square root of 7.. The answer that I got was square root of 14.

I forgot that there is invisible 1 before square root of 7. What I did was I just simply add the 2 values inside the square root sign.

I will always remember that scenario whenever I will encounter square r0ots..

thank you very much for your help! More powers to you! :)

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