Question

How can I solve this limit problem?

Answer 1

\[\lim_{x \rightarrow 0}\frac{ x-1 }{ x^2(x+5) }\]

Answer 2

What techniques do you know so far?
I'd recommend L'Hopital's Rule

Answer 3

You can also try approximating it numerically.

Answer 4

I'm familiar with L'Hopital's Rule but is there any way of finding it through limit laws?

Answer 5

I've expanded the denominator to: \[x^3+5x^2.\]

Answer 6

There are various tricks for rewriting expressions to make the division-by-zero problem go away.

Answer 7

After that I tried dividing everything by the largest degree, in this case 3.

Answer 8

You mean to get something like (x^-2 - x^-3)/(1+5x^-1) ?

Answer 9

Yes, but it doesn't work because I can't really find the limits of x^-3, and x^-1

Answer 10

Right . .
You might try a substitution like let y=1/x, then find limit as y->∞

Answer 11

I hadn't thought of that, thanks for the tip. Hope it works.

Answer 12

Me too! After I learned L'Hopital's Rule I kinda forgot how to do it any other way. :">

Answer 13

I can't seem to solve it using your suggestion.

Answer 14

You are able to see that the limit is zero, though, right?

Answer 15

Yes, where are you heading with that?

Answer 16

To start a reverse-engineering method.

Answer 17

I've tried decomposing the numerator into (sqrtx-1)(sqrtx+1), but I can't come up with anything else.

Answer 18

I'll go check my old calc book to see if there's some other trick.

Answer 19

Maybe we're overthinking this. Seems to me it goes to -1/0, so the limit is -infinity.

Answer 20

Yeah, I just saw that.

Answer 21

The fraction can be broken into (1/x^2)/((x-1)/(x+5)) which would yield infinity*(-1)

Answer 22

Oops, \[\lim_{x \rightarrow 0}\frac{ 1 }{ x^2 }\times \lim_{x \rightarrow 0}\frac{ x-1 }{ x+5 }=\infty \times \frac{ -1 }{ 5 }=- \infty\]

Answer 23

Thanks CliffSedge for going as far as consulting a book to help me solve this problem, and thanks creeksider for pointing its simplicity.

Answer 24

Yeah, I started getting -∞ after isolating the 1/x^2 too, but it contradicted what I got using L'Hopital's Rule. I guess I just used L'Hopital incorrectly.
Looking at the graph, -∞ is obviously correct.

Answer 25

To apply L'Hopital's rule you need something that goes to 0 on top and bottom, or to infinity on top and bottom, so it doesn't apply here.