Help with a limit question

Apr 2008
80
2
Please determine the following limit if they exist. If the limit doe not exist put DNE.
lim 2x^3 / x^2 + 10x - 12
x->infinity.
Thanks.
 
Nov 2009
517
130
Big Red, NY
Please determine the following limit if they exist. If the limit doe not exist put DNE.
lim 2x^3 / x^2 + 10x - 12
x->infinity.
Thanks.
Do you know L'Hospitals?

\(\displaystyle \lim_{x\to\infty} \frac{2x^3}{x^2 + 10x - 12}\)

\(\displaystyle = \lim_{x\to\infty} \frac{6x^2}{2x + 10}\)

\(\displaystyle = \lim_{x\to\infty} \frac{12x}{2} = \lim_{x\to\infty} 6x \to\infty\) \(\displaystyle \fbox{DNE}\)
 
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Aug 2007
3,171
860
USA
You may wish to review your "Order of Operations" and add a few more parentheses, but other than that, with rational functions, degree of numerator greater than degree of denominator ==> Definitely DNE.
 
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skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
Please determine the following limit if they exist. If the limit doe not exist put DNE.
lim 2x^3 / x^2 + 10x - 12
x->infinity.
Thanks.
very straight-forward limit ... what do you think?
 
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Dec 2009
3,120
1,342
Please determine the following limit if they exist. If the limit doe not exist put DNE.
lim 2x^3 / x^2 + 10x - 12
x->infinity.
Thanks.
Hi rowdy3,

As the numerator contains a higher power of x, then the expression
increases without bound as x does, hence the limit DNE.

\(\displaystyle \frac{x^3}{x^2+10x-12}=\frac{\frac{1}{x}\left(x^3\right)}{\frac{1}{x}\left(x^2+10x-12\right)}\)

\(\displaystyle =\frac{x^2}{x+10-\frac{12}{x}}\)

As \(\displaystyle x\ \rightarrow\ \infty\) the term \(\displaystyle \frac{12}{x}\ \rightarrow\ 0\)

\(\displaystyle \frac{\frac{1}{x}\left(x^2\right)}{\frac{1}{x}(x+10)}=\frac{x}{1+\frac{10}{x}}\)

As \(\displaystyle x\ \rightarrow\ \infty\) the term \(\displaystyle \frac{10}{x}\ \rightarrow\ 0\)

\(\displaystyle \lim_{x\ \rightarrow\ \infty}\frac{x}{1}\) DNE

If you like, do it in one step, eliminating x from the denominator by multiplying by \(\displaystyle \frac{\left(\frac{1}{x^2}\right)}{\left(\frac{1}{x^2}\right)}\)
 
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