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Math Help - Trouble finding limits

  1. #1
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    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
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    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..
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    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
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  4. #4
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    Awesome! thanks for the help, i dont think to use L'hopital's rule again
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  5. #5
    MHF Contributor arbolis's Avatar
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    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.
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    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?
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    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}
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  8. #8
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    Hello,

    Quote Originally Posted by Mathstud28 View Post
    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.
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    If you would

    have read my earlier posts I already went over those methods...thanks for helping though!
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  10. #10
    Moo
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    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 ^ ^
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  11. #11
    MHF Contributor arbolis's Avatar
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    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.
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Just wondering

    Quote Originally Posted by Mathstud28 View Post
    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?
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  13. #13
    MHF Contributor arbolis's Avatar
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    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.
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    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
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  15. #15
    Moo
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    Quote Originally Posted by Mathstud28 View Post
    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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