I've been trying to figure this out for a while and I still have no managed to yet...

lim

x-> 1

X^2 + 1 / (Square root of 2X + 2) - 2

On the back of the book it says the answer is 4.

Can anyone tell me how to do this?

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- Jan 13th 2009, 03:59 PMBrazucaIndicate whethere limit exists
I've been trying to figure this out for a while and I still have no managed to yet...

lim

x-> 1

X^2 + 1 / (Square root of 2X + 2) - 2

On the back of the book it says the answer is 4.

Can anyone tell me how to do this? - Jan 13th 2009, 04:06 PMProve It
- Jan 13th 2009, 04:44 PMBrazuca
I mixed up the numbers it is.

X^2 + 1 / (Square root of 2X + 2) - 2

The -2 is part of the denominator so it is in the bottom.

And the + 2 is also inside the square root. - Jan 13th 2009, 04:45 PMProve It
So it's

$\displaystyle \frac{x^2 + 1}{\sqrt{2x + 2} - 2}$?

My CAS is telling me that it's undefined, so I guess, no the limit does not exist.

You could probably show this by multiplying top and bottom by the bottom's conjugate, and then showing that the limit is still not able to be found... - Jan 13th 2009, 06:26 PMBrazuca
Yes thats exactly it.

For some reason the book shows that the answer is 4. The book is "Calculus One and several variables" Tenth Edition.

Here's the question Section 2.1 page 63.

49) Decide on intuitive grounds whther or not the indicated limit exists; evaluate the limit if it does exist.

lim x -> 1

http://www.mathhelpforum.com/math-he...4a0d0829-1.gif

Answer on the back of the book. 4 - Jan 13th 2009, 07:16 PMmr fantastic