Ap calculus: lim (x->infty) 2x^4+6^2+5 / 3+x^3

[joy08]

Find the limit : Lim approaches positive infinity 2x^4+6^2+5/3+x^3.

This is what I figure but i don't think its fully right but since I am taking the limit of the function as x approaches infinity, I will only look at the numerators degree term and the denominators degree term. . So we only consider:

x^4/x^3 which simplifies to x.

that means the limit approaches infinity.

 

[o_O]

Re: Ap calculus limit

That's basically it. Multiply top and bottom by \(\displaystyle \frac{1}{x^{4}}\) keeping in mind that \(\displaystyle \lim_{x \to \infty} \frac{1}{x^{n}} = 0\) where n > 0.

 

[stapel]

Find the limit : Lim approaches positive infinity 2x^4+6^2+5/3+x^3.

What you have posted means the following:

. . . . .lim [sub:3ae8yg7f]x -> infty[/sub:3ae8yg7f] 2x[sup:3ae8yg7f]4[/sup:3ae8yg7f] + 36 + (5/3) + x[sup:3ae8yg7f]3[/sup:3ae8yg7f]

Would it be correct to guess that you actually mean the following?

. . . . .lim [sub:3ae8yg7f]x -> infty[/sub:3ae8yg7f] (2x[sup:3ae8yg7f]4[/sup:3ae8yg7f] + 6x[sup:3ae8yg7f]2[/sup:3ae8yg7f] + 5) / (3 + x[sup:3ae8yg7f]3[/sup:3ae8yg7f])

Or did you mean something else...?

Thank you!

Eliz.

 

[galactus]

Please use proper grouping symbols and make use preview to make sure there are no typos in your problem.

\(\displaystyle \lim_{x\to\infty}\frac{2x^{4}+6x^{2}+5}{x^{3}+3}\)

Take note that the power of the numerator is higher than the power of the the denominator.

You are correct. We can disregard all the lower powers and end up with \(\displaystyle \lim_{x\to\infty}x={\infty}\)

Good work, except, grouping symbols, grouping symbols.