# Thread: when x approaches to -inf in limit

1. ## when x approaches to -inf in limit

Hi,
I'm having problems with few limit questions

lim x-> -inf (x^3 - 1)/[(2x^2)(x-1)]

lim x-> -inf (6x^2 - x +3)/(3-x-4x^2)

lim x-> -inf (sqrt (16x^2 - 5x))/(3x+1)

i have to solve these problems by divide by highest power in denominator thingy

what i am thinking is that:

for the first one, i think i should multiply by -1 for denominator because the highest power is 3 ?

for the second one, i think i should leave it as it is because the highest power is 2 ?

for the third one, i think i should multiply by -1 for denominator because the highest power is 1 ?

am i right? if not, could you point me out where i went wrong ?

2. Originally Posted by haebinpark
Hi,
I'm having problems with few limit questions

lim x-> -inf (x^3 - 1)/[(2x^2)(x-1)]

lim x-> -inf (6x^2 - x +3)/(3-x-4x^2)

lim x-> -inf (sqrt 16x^2 - 5x)/(3x+1)

i have to solve these problems by divide by highest power in denominator thingy

what i am thinking is that:

for the first one, i think i should multiply by -1 for denominator because the highest power is 3 ?

for the second one, i think i should leave it as it is because the highest power is 2 ?

for the third one, i think i should multiply by -1 for denominator because the highest power is 1 ?

am i right? if not, could you point me out where i went wrong ?

There's a trick you can always try: put $t:=-x\Longrightarrow x\rightarrow -\infty \Longleftrightarrow t\rightarrow \infty$ , so for example, and taking into account that $A^3+B^3=(A+B)(A^2-AB+B^2)$:

$\lim_{x\to -\infty}\frac{x^3-1}{2x^2(x-1)}=\lim_{t\to \infty}\frac{-t^3-1}{2t^2(-t-1)}=$ $\lim_{t\to \infty}\frac{t^3+1}{2t^2(t+1)}=\lim_{t\to \infty}\left(\frac{t^2-t+1}{2t^2}\cdot \frac{t^{-2}}{t^{-2}}\right)=\lim_{t\to \infty}\frac{1-1\slash t+1\slash t^2}{2}=\frac{1}{2}$ .

Now you try the other ones by yourself

Tonio

3. Originally Posted by haebinpark
Hi,
I'm having problems with few limit questions

lim x-> -inf (x^3 - 1)/[(2x^2)(x-1)]

lim x-> -inf (6x^2 - x +3)/(3-x-4x^2)

lim x-> -inf (sqrt 16x^2 - 5x)/(3x+1)

i have to solve these problems by divide by highest power in denominator thingy

what i am thinking is that:

for the first one, i think i should multiply by -1 for denominator because the highest power is 3 ?

for the second one, i think i should leave it as it is because the highest power is 2 ?

for the third one, i think i should multiply by -1 for denominator because the highest power is 1 ?

am i right? if not, could you point me out where i went wrong ?
$\lim_{x \to -\infty}{\frac{x^3 - 1}{2x^2(x-1)}}=\lim_{x \to -\infty}{\frac{1 - \frac{1}{x^3}}{2-\frac{1}{x}}}=\frac{1}{2}$
$\lim_{x \to -\infty}{\frac{6x^2 - x +3}{3-x-4x^2}}=\lim_{x \to -\infty}{\frac{6 - \frac{1}{x} +\frac{3}{x^2}}{\frac{3}{x^2}-\frac{1}{x}-4}}=\frac{-3}{2}$
$\lim_{x \to -\infty}{\frac{\sqrt {16x^2} - 5x}{3x+1}}=\lim_{x \to -\infty}{\frac{-4x - 5x}{3x+1}}=\lim_{x \to -\infty}{\frac{-9}{3+\frac{1}{x}}}=-3$

4. so, none of them need to be change the sign in denominator?

5. thank you for the help tonio,
but i don't really understand how it goes like that,
could you explain again in a little easier way?

6. ## common procedure

Originally Posted by lovemath
$\lim_{x \to -\infty}{\frac{x^3 - 1}{2x^2(x-1)}}=\lim_{x \to -\infty}{\frac{1 - \frac{1}{x^3}}{2-\frac{1}{x}}}=\frac{1}{2}$
$\lim_{x \to -\infty}{\frac{6x^2 - x +3}{3-x-4x^2}}=\lim_{x \to -\infty}{\frac{6 - \frac{1}{x} +\frac{3}{x^2}}{\frac{3}{x^2}-\frac{1}{x}-4}}=\frac{-3}{2}$
$\lim_{x \to -\infty}{\frac{\sqrt {16x^2} - 5x}{3x+1}}=\lim_{x \to -\infty}{\frac{-4x - 5x}{3x+1}}=\lim_{x \to -\infty}{\frac{-9}{3+\frac{1}{x}}}=-3$
as lovemath did most elementary problems involving limits to infinity can be solved by dividing by the highest power of x.