limit 3x^7-2x-1/4x^7-9 as x goes to infinity

[Jade]

limit x to infinity

3x^7-2x-1/4x^7-9

I am wanting to make this 3-2x-1/4-9 by cancelling out the numbers going to the 7th power.

How do you plug in infinity? I really need to figure out how to wrap my brain around this?!

 

[galactus]

What is it you want to do?.

\(\displaystyle \L\\\lim_{x\to\infty}\frac{3x^{7}-2x-1}{4x^{7}-9}?\)

 

[Jade]

Evaluate the limit

I need to evaluate the limit x goes to infinity

 

[pka]

\(\displaystyle \L \frac{{3x^7 - 2x - 1}}{{4x^7 - 9}} = \frac{{3 - \frac{2}{{x^6 }} - \frac{1}{{x^7 }}}}{{4 - \frac{9}{{x^7 }}}}\)

 

[Jade]

You divided by x^7

you divided by x^7

=3/4

 

[Jade]

What if there is more than one exponent

limit x goes to negative infinity

3x^4+2/2x-4x^4+4x^5

Would you take the greatest one x^5

 

[pka]

Here is a basic rule of thumb. Like all rules of thumbs there are exceptions.
\(\displaystyle \lim _{x \to \pm \infty } \frac{{g(x)}}{{h(x)}}\) has a finite limit if deg[g(x)]= deg[h(x)].
If has limit 0 if deg[g(x)] < deg[h(x)].
It has no finite limit if deg[g(x)] > deg[h(x)].

But be careful it only a rule of thumb.

 

[Jade]

Limit=0

So in my situation above 3x^4+2/2x-4x^4+4x^5 when x goes to negative infinity you can determine when evaluating the limit that it is zero because the degree of the numerator is less than the degree of the denominator?

 

[tkhunny]

3x^4+2/2x-4x^4+4x^5

No, in this case, there is no limit.

If you meant this, (3x^4+2)/(2x-4x^4+4x^5), then your thumb rule is working OK.