Find the limits of the following functions

[Mattofix]

Homework Statement

Ok – two limit questions. Find the limit as x -> infinity or if no limit exists then prove so.

i) (x + log(x^2))/(3x+2)
ii) x/(1 + (x^2)(sin x)^2

The Attempt at a Solution

i) my first thought was that log(x^2)/x tends to 0 so the function tends to 1/3, but if this is the case do I need to prove that log(x^2)/x tends to 0?

ii) well 0 <= (sin x)^2 <= 1 , I am pretty stuck on this one.

 

[fermio]

Does log here means ln? If so then by using L'hopital rule
$$\lim_{x\to\infty}\frac{x+\ln x^2}{3x+2}=\lim_{x\to\infty}\frac{(x+\ln x^2)'}{(3x+2)'}=\lim_{x\to\infty}\frac{1+\frac{2x}{x^2}}{3}=\frac{1}{3}$$

$$\lim_{x\to\infty}\frac{x}{1+x^2\sin^2 x}=\lim_{x\to\infty}\frac{1}{2x\sin^2 x+x^2 2\sin x\cos x}$$

 

[Mattofix]

thanks - i) makes sense.

so ii) -> 0 ?

 

[VietDao29]

$$\lim_{x\to\infty}\frac{x}{1+x^2\sin^2 x}=\lim_{x\to\infty}\frac{1}{2x\sin^2 x+x^2 2\sin x\cos x}$$

Well, L'Hospital does not work for cases, in which, the limit does not exist.

so ii) -> 0 ?

Nope.

You can think like this: as x tends to infinity, sin(x) can take any value on the interval [-1, 1].

If sin(x) = 0, your expression becomes: x.
And if sin(x) = 1, then your expression will become: x / (1 + x²)

So now, we'll choose 2 sequences (namely, x_n, and x'_n), both of which grow without bounds, and sin(x_n) = 0, sin(x'_n) = 1, for all n.

We can choose x_n = 2n(pi), and x'_n = pi/2 + 2n(pi).
The 2 sequences above satisfy all requirements above (you can check it yourself).

What can you say about the 2 limits:

$$\lim_{n \rightarrow \infty} \frac{x_n}{1 + x_n ^2 \sin ^ 2 (x_n)} , \quad n \in \mathbb{N}$$

and:

$$\lim_{n \rightarrow \infty} \frac{x'_n}{1 + x'_n ^2 \sin ^ 2 (x'_n)} , \quad n \in \mathbb{N}$$

From there, what can you conclude about the limit:

$$\lim_{x \rightarrow \infty} \frac{x}{1 + x ^2 \sin ^ 2 (x)}$$?

Can you go from here? :)

 

[Mattofix]

yep - thanks