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 January 9th, 2011, 08:37 AM #1 Newbie   Joined: Jan 2011 Posts: 3 Thanks: 0 Limit to Infinity Hi, Next week I have a calculus exam, so I tried some old exams. There is one question I just cant solve: Limit[(2x^2-6^x)/((3x+2^x)(3^x-x)),x->infinity] I used mathematica to get the answer so I know it is 1 but I have no idea why... Can anyone explain it to me
 January 9th, 2011, 08:58 AM #2 Newbie   Joined: Jan 2011 Posts: 22 Thanks: 0 Re: Limit to Infinity
January 9th, 2011, 09:08 AM   #3
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Re: Limit to Infinity

Quote:
 Originally Posted by Chimoo http://en.wikipedia.org/wiki/Limit_(mathematics)
What should I do with that... I have a calculus book with a lot more explanation than that and I tried sosmath.com but I just cant find how to do it.

 January 9th, 2011, 09:20 AM #4 Newbie   Joined: Jan 2011 Posts: 22 Thanks: 0 Re: Limit to Infinity Sorry if that link wasn't very helpful, try this one http://www.calculus-help.com/how-do-...aluate-limits/
 January 9th, 2011, 09:25 AM #5 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Re: Limit to Infinity Hello Stef, Or: $\lim_{x \to \infty} \frac{2x^2-6^x}{(3x+2^x)(3^x-x)}= \\ \lim_{x \to \infty} \frac{2x^2-6^x}{6^x-3 x^2-x \cdot 2^x +x \cdot 3^{x+1}}= \\ \\ \lim_{x \to \infty} \frac{2x^2-6^x+\left( x^2+x \cdot 2^x-x \cdot 3^{x+1}\right)}{6^x-3 x^2-x \cdot 2^x +x \cdot 3^{x+1}}-\frac{\left( x^2+x \cdot 2^x-x \cdot 3^{x+1}\right)}{6^x-3 x^2-x \cdot 2^x +x \cdot 3^{x+1}}= \\ \\ \lim_{x \to \infty}-1 + \lim_{x \to \infty} \frac{\left( x^2+x \cdot 2^x-x \cdot 3^{x+1}\right)}{6^x-3 x^2-x \cdot 2^x +x \cdot 3^{x+1}}=-1+0=-1$ Because $\lim_{x \to \infty} \frac{3^{x+1}}{6^x}=\lim_{x \to \infty} 3^1 \cdot \frac{3^{x}}{6^x}=3 \lim_{x \to \infty} 0.5^x=3 \cdot 0=0$. Same calculations can be written included the other terms in the numerator of the limit (which is $\left( x^2+x \cdot 2^x-x \cdot 3^{x+1}\right)$) Got it? Hoempa
 January 9th, 2011, 10:25 AM #6 Senior Member   Joined: Nov 2010 From: Staten Island, NY Posts: 152 Thanks: 0 Re: Limit to Infinity Let me give you a quick shortcut: Just take the term with the fastest growing function on top and divide by the term with the fastest growing function on bottom. Then take the limit. This is $\frac{-6^x}{6^x}=-1$ Remarks: (1) We needed to multiply out the denominator in order to find the fastest growing term. (2) An exponential function with base greater than 1 grows faster than any polynomial function.
January 9th, 2011, 01:04 PM   #7
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Re: Limit to Infinity

Quote:
 Originally Posted by DrSteve Let me give you a quick shortcut: Just take the term with the fastest growing function on top and divide by the term with the fastest growing function on bottom. Then take the limit. This is $\frac{-6^x}{6^x}=-1$ Remarks: (1) We needed to multiply out the denominator in order to find the fastest growing term. (2) An exponential function with base greater than 1 grows faster than any polynomial function.
So if you have a complex limit like this you just look for the highest exponential term or otherwise the highest polynomial term in both the numerator and the denominator (or however you call them in English) and you look what is the fast growing one... well this is the best explanation for me.
Maybe more questions will follow for this exam, is kinda sucks...
Thanks
*Solved*

 January 9th, 2011, 04:43 PM #8 Senior Member   Joined: Nov 2010 From: Staten Island, NY Posts: 152 Thanks: 0 Re: Limit to Infinity Yes. That is essentially correct as long as those are the only types of terms that appear. This will also work if there are logarithms and/or roots. Keep in mind that exponential functions grow faster than polynomials which grow faster than logarithmic functions. Also a function with a root can be treated like a polynomial by using fractional exponents. The idea is that as x gets larger the faster growing function takes over, and all the slower growing ones become negligible. So "in the limit" the slower growing ones disappear completely (this is of course a very informal argument). I would recommend that you make sure that you are allowed to use these shortcuts on your exams. I think that most teachers are ok with them (they always give the correct answer after all), but some may want you to show the details in which case you'll only get partial credit.

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