Using epsilon-N definition to prove limits

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October 19th 2009, 08:21 AM
leftfootwonder7

Using epsilon-N definition to prove limits

Hi,
im stuck with a question on proving a limit of a sequence. Here it is:

Use the epsilon-N definition of a limit to prove that:

lim(n>infinity) n+5/n^2 + 2 = 0.

For epsilon = 1/25 give N such that for all n>=N it follows that:

|n+5/n^2 + 2| < epsilon.

Not sure how to start really. Any help would be appreciated. Thanks
October 19th 2009, 08:55 AM
tonio

Quote:

Originally Posted by leftfootwonder7

Hi,
im stuck with a question on proving a limit of a sequence. Here it is:

Use the epsilon-N definition of a limit to prove that:

lim(n>infinity) n+5/n^2 + 2 = 0.

For epsilon = 1/25 give N such that for all n>=N it follows that:

|n+5/n^2 + 2| < epsilon.

Not sure how to start really. Any help would be appreciated. Thanks

Try to add parentheses to make clear to others your expressions.

Let $\epsilon > 0$ be arbitrary. We want to show that for all but a finite number of indexes $n$ we have

$\mid\frac{n+5}{n^2+2}\mid<\epsilon$

As the expression between the absolute value is always positive we can drop the abs. value, so cross multiplying we get:

$n+5 <\epsilon n^2 + 2\epsilon \Longleftrightarrow \epsilon n^2-n+2\epsilon -5 >0$

Wel, now solve the quadratic inequality above and find your n!

Tonio
October 29th 2009, 04:05 PM
Roam

Quote:

Originally Posted by tonio

Try to add parentheses to make clear to others your expressions.

Let $\epsilon > 0$ be arbitrary. We want to show that for all but a finite number of indexes $n$ we have

$\mid\frac{n+5}{n^2+2}\mid<\epsilon$

As the expression between the absolute value is always positive we can drop the abs. value, so cross multiplying we get:

$n+5 <\epsilon n^2 + 2\epsilon \Longleftrightarrow \epsilon n^2-n+2\epsilon -5 >0$

Wel, now solve the quadratic inequality above and find your n!

Tonio

Hi Tonio, I don't understand why we couldn't just use the fact that:

$<br /> \frac{n+5}{n^2+2}< \frac{n+5}{n^2} = \frac{6}{n} < \epsilon<br />$

And then we can take $N= \frac{6}{\epsilon}$

??
October 29th 2009, 07:26 PM
tonio

Quote:

Originally Posted by Roam

Hi Tonio, I don't understand why we couldn't just use the fact that:

$<br /> \frac{n+5}{n^2+2}< \frac{n+5}{n^2} = \frac{6}{n} < \epsilon<br />$

And then we can take $N= \frac{6}{\epsilon}$

??

Who says we couldn't? Of course we could, it's just that you saw it and I didn't.(Wink)

Of course, you meant above $\frac{5+n}{n^2} \leq \frac{6}{n}$ , and not equality.

Well done...and you answered your own question.

Tonio