View Full Version : Topic is about Limits
thinay
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
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})}{(\s qrt{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
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
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
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! :)