# Thread: How to calculate the following limit

1. ## How to calculate the following limit

I'm struggling a bit with this one. Can someone please show me the work required to find the result. I know the answer is $\frac{7}{16}$ but I'm not sure how to arrive at that.

The question is:

$\lim \frac{-7x^2-4x+16}{-16x^2+3x+3} as x \rightarrow -\infty$

2. Originally Posted by softwareguy
I'm struggling a bit with this one. Can someone please show me the work required to find the result. I know the answer is $\frac{7}{16}$ but I'm not sure how to arrive at that.

The question is:

$\lim \frac{-7x^2-4x+16}{-16x^2+3x+3} as x \rightarrow -\infty$
You can use L'Hopital's rule here.

3. Originally Posted by ur5pointos2slo
You can use L'Hopital's rule here.
I shouldn't assume you have learned L'Hopital's rule yet. Let me show you.

lim x-> - infinity (-7x^2-4x+16) / (-16x^2 + 3x + 3)

Plugging in infinity you get infinity/infinity which is L'Hop material

so take the derivative of the top and bottom

lim x-> - infinity (-14x-4) /(-32x+3) again infinity/infinity

take the derivative of the top and bottom again

lim x-> - infinity (-14/-32) = 7/16

4. $\lim_{x\to -\infty} \frac{-7x^2-4x+16}{-16x^2+3x+3} = \lim_{x\to -\infty} \frac{-\frac{7x^2}{x^2}-\frac{4x}{x^2}+\frac{16}{x^2}}{-\frac{16x^2}{x^2}+\frac{3x}{x^2}+\frac{3}{x^2}} = \lim_{x\to -\infty} \frac{-7-\frac{4}{x}+\frac{16}{x^2}}{-16+\frac{3}{x}+\frac{3}{x^2}}$ $= \frac{-7-0+0}{-16+0+0} = \frac{-7}{-16} = \frac{7}{16}

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