f has an infinite limit at infinity proofing

[tsikitsaki]

Hi! How to prove this one?

I tried using epsilon and such, but i get stuck.

 

[JeffM]

[MATH]- 1 \le cos(x) \le 1 \text { for any x.}[/MATH]
[MATH]1 < x \implies 3 < 3x \implies 0 < 2 < 3x + cos(x) \le 3x + 1 < 4x \implies 0 < \dfrac{1}{4x} < \dfrac{1}{3x + cos(x)}.[/MATH]
[MATH]1 < x \implies 2 < x^2 + x \implies 0 < x^2 + x \implies \dfrac{x^2 + x}{4x} < \dfrac{x^2 + x}{3x + cos(x)} \implies \dfrac{1}{4} (x + 1) < \dfrac{x^2 + x}{3x + cos(x)}.[/MATH]
What is [MATH]\lim_{x \rightarrow \infty} \dfrac{1}{4} (x + 1)?[/MATH]
Intuitively, what does that tell you about

[MATH]\lim_{x \rightarrow \infty} \dfrac{x^2 + x}{3x + cos(x)}?[/MATH]
How can you use the formal definitions to bridge the gap between intuition and proof?

 

[lookagain]

Hi! How to prove this one?
View attachment 25780

You might consider an additional approach:



As \(\displaystyle \ -1 \le cos(x) \le 1, \ \ \displaystyle\lim_{x \to \infty} \dfrac{cos(x)}{x} \ = \ 0.\)

\(\displaystyle \displaystyle\lim_{x \to \infty} \dfrac{x^2 \ + \ \ x \ \ }{ \ 3x \ + \ cos(x) \ } \ = \)

\(\displaystyle \displaystyle\lim_{x \to \infty} \dfrac{\dfrac{x^2}{x} \ \ \ \ + \ \ \ \dfrac{x}{x} \ \ }{\dfrac{3x}{x} \ + \ \dfrac{cos(x)}{x}} \ =\)

\(\displaystyle \displaystyle\lim_{x \to \infty} \dfrac{ \ x \ + \ 1 \ }{ \ 3 \ + \ 0 \ } \ =\)

\(\displaystyle \displaystyle\lim_{x \to \infty}\dfrac{1}{3}(x + 1) \ =\)

 

[lex]

Or using L'Hôpital's rule: (with justifications)

[MATH]\lim_{x\rightarrow\infty}{\left(\frac{x^2+x}{3x+cos\ x}\right)}[/MATH]
[MATH]\lim_{x\rightarrow\infty}{\left(\frac{2x+1}{3-sin\ x}\right)}[/MATH]
=[MATH]\infty[/MATH]
since [MATH]3-sin\ x[/MATH] is bounded: [MATH]2\le3-sinx\le4[/MATH]