# Thread: Help with a limit question

1. ## Help with a limit question

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.

2. Originally Posted by rowdy3
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?

$\lim_{x\to\infty} \frac{2x^3}{x^2 + 10x - 12}$

$= \lim_{x\to\infty} \frac{6x^2}{2x + 10}$

$= \lim_{x\to\infty} \frac{12x}{2} = \lim_{x\to\infty} 6x \to\infty$ $\fbox{DNE}$

3. 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.

4. Originally Posted by rowdy3
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?

5. Originally Posted by rowdy3
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.

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

$=\frac{x^2}{x+10-\frac{12}{x}}$

As $x\ \rightarrow\ \infty$ the term $\frac{12}{x}\ \rightarrow\ 0$

$\frac{\frac{1}{x}\left(x^2\right)}{\frac{1}{x}(x+1 0)}=\frac{x}{1+\frac{10}{x}}$

As $x\ \rightarrow\ \infty$ the term $\frac{10}{x}\ \rightarrow\ 0$

$\lim_{x\ \rightarrow\ \infty}\frac{x}{1}$ DNE

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