Trouble finding limits

**cowboys111** · April 6th 2008, 01:26 PM

Hi, Im having trouble finding the limit as n goes to infiniti of ((n+1)(n+2))/(2n^2). Can someone please help me start this problem?
Thank you

**Mathstud28** · April 6th 2008, 01:33 PM

$\lim_{n \to {\infty}}\frac{(n+1)(n+2)}{2n^2}=lim_{n \to {\infty}}\frac{n^2+3n+2}{2n^2}$ ...now direct substitution yields $\frac{0}{0}$ ...so you can use L'hopital's rule so $\lim_{n \to {\infty}}\frac{(n+1)(n+2)}{2n^2}=\lim_{n \to {\infty}}\frac{2n+3}{4n}$ ...once again the indeterminate form $\frac{0}{0}$ ..so L'hopitals rule again so $\lim_{n \to {\infty}}\frac{(n+1)(n+2)}{2n^2}=\lim_{n \to {\infty}}\frac{2}{4}=\frac{1}{2}$ ..or if its polynomial with same highest degree exponents you jsut divide the coefficents..

**Mathstud28** · April 6th 2008, 01:34 PM

remember that if the highest degree variable is in the numerator and it is a polynomial the limit is ∞ and if the highest degree variable is in the denominator with polynomials the limit is 0

**cowboys111** · April 6th 2008, 02:00 PM

Awesome! thanks for the help, i dont think to use L'hopital's rule again

**arbolis** · April 7th 2008, 08:59 AM

Your expression is (n+1)(n+2)/2n^2. It's equal to (n^2+3n+2)/2n^2. When you have this, factorize by the variable of the major exponent. In this case it's n^2. So we get (n^2(1+3/n+2/n^2)/(n^2*(2)). Simplify the numerator with the denominator. We get (1+3/n+2/n^2)/2. As n tends to positive infinite, the expression tends to 1/2.
This is usally the method teached before to learn l'Hôpital rule.

**Mathstud28** · April 7th 2008, 09:02 AM

now my question for you is...is that really easier than using L'hopital's rule or regognizing the equal order variables in the numerator and denominator?

**Mathstud28** · April 7th 2008, 09:14 AM

if this is what you did arbolis I am sorry I am just making it clear through LaTeX...ok you have $\lim_{n \to {\infty}}\frac{(n+1)(n+2)}{2n^2}=\lim_{n \to {\infty}}\frac{n^2+3n+2}{2n^2}=\lim_{n \to {\infty}}\frac{n^2+3n+2}{2n^2}\cdot\frac{\frac{1}{ n^2}}{\frac{1}{n^2}}$ which equals $\lim_{n \to {\infty}}\frac{1+\frac{3}{n}+\frac{2}{n}}{2}=\frac {1+0+0}{2}=\frac{1}{2}$

**Moo** · April 7th 2008, 09:18 AM

Hello,

Originally Posted by Mathstud28

now my question for you is...is that really easier than using L'hopital's rule or regognizing the equal order variables in the numerator and denominator?

I do think it's easier. This is a method learnt before l'Hôpital's rule. The study of the degree can help :

limit at infinity :

if deg(numerator)>deg(denominator) -> tends to infinity (sign is to determine with the coefficient of the highest power)

if deg(denominator)>deg(numerator) -> tends to 0

if deg(denominator)=deg(numerator) -> tends to the quotient of the two coefficients corresponding to the highest powers.

**Mathstud28** · April 7th 2008, 09:22 AM

have read my earlier posts I already went over those methods...thanks for helping though!

**Moo** · April 7th 2008, 09:23 AM

Your first message was about using l'Hôpital's rule o.O
And you were wondering if this was easier or not...
Make your mind ^ ^

**arbolis** · April 7th 2008, 10:18 AM

Studying the degree is easier for me, since (sometimes, like in this example) you can solve the limit without requiring to derivate. Just factorizing and simplifying.

**Mathstud28** · April 7th 2008, 10:19 AM

Originally Posted by Mathstud28

if this is what you did arbolis I am sorry I am just making it clear through LaTeX...ok you have $\lim_{n \to {\infty}}\frac{(n+1)(n+2)}{2n^2}=\lim_{n \to {\infty}}\frac{n^2+3n+2}{2n^2}=\lim_{n \to {\infty}}\frac{n^2+3n+2}{2n^2}\cdot\frac{\frac{1}{ n^2}}{\frac{1}{n^2}}$ which equals $\lim_{n \to {\infty}}\frac{1+\frac{3}{n}+\frac{2}{n}}{2}=\frac {1+0+0}{2}=\frac{1}{2}$

but do you basically do this but in your head or do you just think about the coefficients being divided?

**arbolis** · April 7th 2008, 10:38 AM

First I look the expression to look if there is any indetermination. If there is one like in our example, mindly I do the quotient of the variable afected by the major degree taking in count the coefficients. If I have to answer on paper, I do all the steps.

**Mathstud28** · April 7th 2008, 10:43 AM

but what if you were setting up the formal proof that the derivative of $sin(x)$ is $cos(x)$ and you ran into $\lim{\Delta{x} \to 0}\frac{1-cos(\Delta{x})}{\Delta{x}}$ or if you had $\lim_{t \to 0}\frac{\sqrt{1+t^2}-t}{t}$ what then...do you use L'hopitals or do you do something differnet...jus curious

**Moo** · April 7th 2008, 11:09 AM

Originally Posted by Mathstud28

but what if you were setting up the formal proof that the derivative of $sin(x)$ is $cos(x)$ and you ran into $\lim{\Delta{x} \to 0}\frac{1-cos(\Delta{x})}{\Delta{x}}$ or if you had $\lim_{t \to 0}\frac{\sqrt{1+t^2}-t}{t}$ what then...do you use L'hopitals or do you do something differnet...jus curious

Well, i don't think the second one is indeterminated... Limit is + or - infinity depending on 0+ or 0-

You can survive without l'Hôpital's rule

i mean, for basic limits

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