# Thread: negative infinity limits

1. ## negative infinity limits

well the question is: (sorry don't know how to type math)

lim 2x+3
x->-∞ 5x+7

well the answer key says 2/5 but i get -2/5... i looked up online which told me to get rid of the negative by replacing it with u so i did that and still got the same answer.

This is what I did...

lim -2(-x)+3
x->-∞ -5(-x)+7

- lim -2u+3
u->∞ -5u+7

- lim u(-2 + (3/u))
u->∞ u(-5 + (7/u))

-u(-2)
u(-5)

-2/5

this is from instructions online what i normally do is just :

lim 2x+3
x->-∞ 5x+7

x(2 + (3/-∞)
)
x(5 + (7/-∞))

and this gets me like ∞/∞????

I don't get it what am I doing wrong???

2. Adding that "u" is not needed. In your equation you have:

$\lim_{x \to -\infty}\frac{2x+3}{5x+7}
$

From this we want to factor out the highest power of X that occurs in the denominator, and we are left with:

$\lim_{x \to -\infty}\frac{2+\frac{3}{x}}{5+\frac{7}{x}}$

From here, we know that 1/x approaches 0 as x approaches infinity in either direction. And thus we are left with:

$\lim_{x \to -\infty}\frac{2}{5}$

Which is just 2/5.

im -2(-x)+3
x->-∞ -5(-x)+7
This part here, is where you start to confuse yourself. Why are there negatives added on to the 2 and the 5?

- lim u(-2 + (3/u))
u->∞ u(-5 + (7/u))
This is fine (well not really since it's got those negatives but the process is the same), but then you make a mistake here:

-u(-2)
u(-5)
The u's cancel out, but you have made one u negative and the other positive, so they cancel but leave over a -1. Just little errors that are making big problems.

3. Do you know what "L'Hôpital's method" is? You could have used it here since both $2x+3$ and $5x+7$ approach $- \infty$ as x approaches $- \infty$:

$\lim_{x \to -\infty}\frac{2x+3}{5x+7} = \lim_{x \to -\infty}\frac{2}{5} = \frac{2}{5}$

4. Originally Posted by Roam
Actually, you could have used L'Hôpital's method here since both $2x+3$ and $5x+7$ approach $- \infty$ as x approaches $- \infty$:

$\lim_{x \to -\infty}\frac{2x+3}{5x+7} = \lim_{x \to -\infty}\frac{2}{5} = \frac{2}{5}$
It is preferable to avoid L'Hopital's rule when possible on MHF. We do not know at what stage of the original posters course this question originates. They may not have yet done derivatives let alone L'Hopital's rule.

CB

5. Originally Posted by Yello
well the question is: (sorry don't know how to type math)

lim 2x+3
x->-∞ 5x+7

well the answer key says 2/5 but i get -2/5... i looked up online which told me to get rid of the negative by replacing it with u so i did that and still got the same answer.

This is what I did...

lim -2(-x)+3
x->-∞ -5(-x)+7

- lim -2u+3
u->∞ -5u+7
Here is your error. Where did that leading "-" come from?

- lim u(-2 + (3/u))
u->∞ u(-5 + (7/u))

-u(-2)
u(-5)

-2/5

this is from instructions online what i normally do is just :

lim 2x+3
x->-∞ 5x+7

x(2 + (3/-∞)
)
x(5 + (7/-∞))

and this gets me like ∞/∞????

I don't get it what am I doing wrong???