limit as x approaches negative infinity

**student12** · April 5th 2009, 02:42 PM

limit as x -> -inf.
3x^2 + 9
---------
x-5

now, if i divide all of this by the dominant power in the den. then I end up with 3x, which leads me to beleive that the answer is -inf., and it is. but how do I show this? 3(-inf.)= -inf.? am I solving it with the wrong method?

**scherz0** · April 5th 2009, 03:09 PM

Hello there,

$\lim_{x \to -\infty} \frac{3x^2 + 9}{x - 5}$

Divide the function by the highest power present, $x^2$ :

$\lim_{x \to -\infty} \frac{3 + \frac{9}{x^2}}{\frac{x}{x^2} - \frac{5}{x^2}}$

= $\lim_{x \to -\infty} \frac{3 + \frac{9}{x^2}}{\frac{1}{x^2}(x - 5)}$

Although $x \rightarrow -\infty$ gives you 0 as the denominator, you can see that the 0 is approaching from the left, because $\frac{1}{x^2} > 0$ and $(x - 5) < 0$ . Therefore, since you will always have a negative denominator when x approaches negative infinity, your limit is negative infinity (which is said to be inexistent).

In other words, your limit is $\frac{3}{0^-}$ = $-\infty \Rightarrow$ No limit.

I hope that this helps.

**student12** · April 5th 2009, 04:28 PM

I don't really understand how you know whether the zero is being approached from the left or right, but I see that there's probably some way of telling. Anyway, I've just started taking calc, so I'm guessing my answer shouldn't be as complex as there being no limit at all. The answer is -inf. We've never talked about -inf. not being a limit, so I think I can rule that out.

Is it legal for me to divide everything by the biggest power present?
Because then, as scherz0 has found, 3/0 would be my result. I technically could do a sign analysis but that doesn't really work...

So...?

**scherz0** · April 5th 2009, 04:45 PM

Hello,

Thanks for getting back to me. Yes, you are allowed to divide by the largest power present.

Basically, if you have calculated any limit to be either infinity or -infinity, the limit is said not to exist. This is because when you evaluate a limit, you can trying to see where the function is going (its y-coordinate) for a specific x-coordinate. Eivdently, if you have a limit of infinity, there's no finite point which the function is advancing towards so there's no limit.

I will assume that you understand why the denominator is zero. Since the function is heading to negative infinity, the "sign" of the zero must be determined because it can yield meaning to the limit. In the denominator, we have $\frac{1}{x^2} -- (1)$ multiplied by $(x - 5) --(2)$ .

(1) will always be positive, (except x = 0). Since the function is approaching negative infinity, if you plug in a really small negative number, you will still get something +.

(2) When the function approaches negative infinity, (x - 5) will be negative.

Since the denominator = (1)(2) = +(-) = -, that zero is approaching from the left (the negative part of the x-axis). Therefore, you have $\frac{3}{0^-}$ .

I hope that this helps.

**student12** · April 5th 2009, 06:44 PM

Thanks for replying, scherz0 (german name?

I was wondering if I could possibly do a sign analysis on this? I doubt it but maybe there's something I don't know.

Oh, and the way you wrote it in your first reply, is that how I would mathematically show my work?

Anyway, thank you very much for all your help!

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