Thread: Indicate 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?

Originally Posted by Brazuca

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 + x) - 2

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

Can anyone tell me how to do this?

Is that

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.

Originally Posted by Brazuca

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.

So it's

?

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...

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



Answer on the back of the book. 4

Originally Posted by Brazuca

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



Answer on the back of the book. 4

There is a typo in the question. It should read .