Associated Topics || Dr. Math Home || Search Dr. Math

### Finding a Limit

```
Date: 02/28/2002 at 22:47:04
From: Corrie
Subject: Limits

My calculus teacher gave my class the answer to this limit; however,
we cannot figure out how to get the answer that is given. If you could
help me with a proof, I would be very grateful.

1 - sqrt(2x^2 - 1)
lim ------------------
x->1      x - 1

Thanks!
```

```
Date: 03/01/2002 at 08:47:05
From: Doctor Peterson
Subject: Re: Limits

Hi, Corrie.

If we just replace x with 1, we get (1-1)/(1-1), which we can't
evaluate; so we'd like to rearrange the expression so we can cancel
something out. The big thing that is in our way is the "1-sqrt" in the
numerator; how about if we "rationalize the numerator" by multiplying
by the conjugate? Then (for x not equal to 1):

1 - sqrt(2x^2 - 1)   1 + sqrt(2x^2 - 1)
------------------ * ------------------ =
x - 1          1 + sqrt(2x^2 - 1)

1 - (2x^2 - 1)                 2(1 - x^2)
--------------------------- = ---------------------------
(x - 1)(1 + sqrt(2x^2 - 1))   (x - 1)(1 + sqrt(2x^2 - 1))

Can you see the next step? Factor the numerator, cancel, and you will
have an expression that can be evaluated at x = 0, and which is equal
to the original expression for all other x, so that its value at x = 0
is the limit you want.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search