Question

How could I find calculate this limit?
https://i.imgsafe.org/464cf2b23a.jpg

Answer 1

When you're taking the limit as x approaches infinity of a polynomial over a polynomial, keep the following in mind:

If the polynomial in the numerator has a higher degree than the polynomial in the denominator, then the limit as it approaches infinity will be \(\pm \infty\)

If the polynomial in the numerator has the same degree of the polynomial in the denominator, then the limit as it approaches infinity will be \(\Large\sf \frac{coefficient ~of ~the~ polynomial~ in ~the~ numerator}{coefficient ~of~ the ~polynomial ~of~ the ~denominator}\)

If the polynomial in the numerator has a lower degree than the polynomial in the denominator, then the limit as it approaches infinity will be 0.

Let me do a few examples for you:
And I'll keep them simple, and to make the point clear, I will use polynomials that can be simplified so you can see it.

\[\lim_{x\rightarrow \infty} \frac{5x^2 + 10x}{x+2} = \lim_{x\rightarrow \infty} \frac{\frac{5x^2 + 10x}{x}}{\frac{x+2}{x}} \] \[\frac{\lim_{x\rightarrow \infty} 5x + 10}{\lim_{x\rightarrow \infty} 1 + \frac{2}{x}}\] And as x goes to infinity, that means 2/x will got to 0, because as x gets larger and larger, that means 2 over a very huge number is very very close to 0. So we are left with \[\lim_{x\rightarrow \infty} 5x + 10\] and as x approaches infinity, we know that 5x + 10 will also approach infinity. The 10 is so negligible because 5x will approach very huge numbers.

Now another way to look at the same problem.
\[\lim_{x\rightarrow \infty} \frac{5x^2 + 10x}{x+2} = \lim_{x\rightarrow \infty}\frac{5x(x + 2)}{x+2} = \lim_{x\rightarrow \infty} 5x\] And we know that as x approaches infinity, 5x will become 5 times infinity, and will be a very huge and unbounded.

Likewise we can show that
\[\lim_{x\rightarrow \infty} \frac{5x }{3x} = \lim_{x\rightarrow \infty}\frac{\frac{5x}{x}}{\frac{3x}{x}} = \lim_{x\rightarrow \infty} \frac{5}{3} \] And the limit of a contant is the constant. And we can see that it would be the coefficient of the numerator over the coefficient of the denominator.

And finally
\[\lim_{x\rightarrow \infty} \frac{1}{x}\] Using the same logic that we used in the first example, as x becomes very large, for example 1,000,000 then 1/(1,000,000) will be very close to 0. And so as x becomes larger and larger, the limit would approach 0.

Answer 2

I hope that was beneficial, if you have any questions, I'll be glad to assist. :)

Answer 3

Actually, the limit is \( -\infty\)

Answer 4

Near infinity a polynomial is equivalent to the term of greatest exponent. Hence near infinity, one can write
\[
\frac{7 x^3-3 x^2+3 x}{-8 x^2+4 x+3}\equiv -\frac{7 x^3}{8 x^2}\equiv -\frac{7 x}{8}\to -\infty
\]
when \( x\to \infty\)

Answer 5

I agree with @eliesaab ... it's -infinity

Answer 6

@IrishBoy123
Yes, we know that it is the limit as x approaches infinity, but what @eliesaab is mentioning is that the limit will approach negative infinity as x approaches infinity.

Answer 7

i was merely putting the graphic on screen, as I have the Chrome extension

:)

Not commenting on anything material !

Answer 8

Oh, my bad :)

Answer 9

😀

Answer 10

can be easily obtain using l'hopital's rule

Answer 11

Because the degree of the numerator is 1 greater than that of the denominator, actually dividing the numerator by the denom algebraically will produce a quotient which begins with a linear function (in the form y=mx+b). All following terms will go to zero as x approaches infinity. What you're getting here is a "slant asymptote."

Please try this long division and see what you get.

Answer 12

If you let x increase without bound, then yes, y will approach negative infinity.

Just a different way of looking at this situation.

Answer 13

i agree with el