1. ## Trouble finding limits

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

2. ## Sure

$\lim_{n \to {\infty}}\frac{(n+1)(n+2)}{2n^2}=lim_{n \to {\infty}}\frac{n^2+3n+2}{2n^2}$...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..

3. ## Also

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

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

5. ## Doing it without l'Hôpital's rule.

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.

6. ## Haha

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?

7. ## And here is another way of doing it

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 $\lim_{n \to {\infty}}\frac{1+\frac{3}{n}+\frac{2}{n}}{2}=\frac {1+0+0}{2}=\frac{1}{2}$

8. 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.

9. ## If you would

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

And you were wondering if this was easier or not...

11. 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.

12. ## Just wondering

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?

13. 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.

14. ## Ahh

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

15. 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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