How can the limit of sqrt(n^2+n)-n be proven using an epsilon-delta definition?

[yskim19]

Homework Statement

Prove (using epsilon-delta definition only) that the limit of the following expression:
sqrt(n^2+n)-n
is 1/2, as n tends towards infinity.

Homework Equations

For any ε > 0, there exists some natural number N, such that:
n > N gives|f(n) - L| < ε

The Attempt at a Solution

Multiply by the conjugate pair and simplify to obtain:
n/[sqrt(n^2+n)+n]

Taking the expression to be compared against ε
|n/[sqrt(n^2+n)+n] - 1/2| and finding a common denominator gives:

|2n/2[sqrt(n^2+n)+n] - [sqrt(n^2+n)+n]/2[sqrt(n^2+n)+n]|
|[n-sqrt(n^2+n)] / 2[sqrt(n^2+n)+n]|

The numerator is always negative and the denominator is always positive, so the expression can be re-written as:
[sqrt(n^2+n) - n] / 2[sqrt(n^2+n)+n]

I chose to split this up into two terms:
sqrt(n^2+n) / 2[sqrt(n^2+n)+n] minus
n / 2[sqrt(n^2+n)+n]

I have an expression in the form a/b-c/d. Decreasing "b", decreasing "c" and increasing "d" both increase the value of the expression, thus the expression below is strictly greater than the
expression above.

sqrt(n^2+n) / 2[sqrt(n^2+n)] minus
1 / 2[sqrt(n^2+n^2)+n]

This simplifies to 1/2 - 1 / n(2sqrt(2)+1)]

Both terms in this expression are positive. Thus, changing the subtraction to addition will strictly increase the value of the entire quantity.

Thus, comparing
1/2 + 1 / n(2sqrt(2)+1) against ε is enough.

Solving for n results in some mess that results in:
n > (positive constant) / (ε - 1/2)

Clearly, something has gone awry. Consideration of an arbitrarily small ε reveals the RHS to be negative. That would imply that the first term in my sequence is already arbitrarily close to my limit.

Where did I go wrong?

 

[Dick]

I'm not sure. Why don't you just take your result n/(sqrt(n^2+1)+n) and write it as n/(n*sqrt(1+1/n^2)+n), cancel the n's and give it a fresh thought.

 

[Mark44]

Here's an approach that's simpler, that I think you can use.
You can start with this inequality:
\sqrt{n^2 + n} - n &lt; \epsilon

You don't need absolute values, since the radical expression is always > n for n > 0, hence the expression on the left side above is always positive.

Now, add n to both sides, and then square both sides, the idea being that if a < b, then a^2 < b^2, where a and b are positive.

That gets you to
n^2 + n &lt; (n + \epsilon)^2

After cleaning a bit, you get to
\epsilon^2 + 2n\epsilon - n &gt; 0

That's a quadratic in epsilon, so that should get you a relationship between epsilon and n, from which you can find the number N that you want.

 

[Dick]

Here's an approach that's simpler, that I think you can use.
You can start with this inequality:
\sqrt{n^2 + n} - n &lt; \epsilon

You don't need absolute values, since the radical expression is always > n for n > 0, hence the expression on the left side above is always positive.

Now, add n to both sides, and then square both sides, the idea being that if a < b, then a^2 < b^2, where a and b are positive.

That gets you to
n^2 + n &lt; (n + \epsilon)^2


After cleaning a bit, you get to
\epsilon^2 + 2n\epsilon - n &gt; 0

That's a quadratic in epsilon, so that should get you a relationship between epsilon and n, from which you can find the number N that you want.

Uh, you don't want to show |sqrt(n^2+n)-n|<e. It isn't. The limit is 1/2. You want to show |sqrt(n^+n)-n-1/2|<e.

 

[Mark44]

Oops! Well, no wonder it was easier. After my first stab got nowhere, I used a clean sheet of paper and omitted that 1/2.