# Find the limits of the following functions

• Mattofix
In summary: The limit of the first sequence is:\lim_{n \rightarrow \infty} \frac{2n(pi)}{1+2n(pi)} and the limit of the second sequence is:\lim_{n \rightarrow \infty} \frac{1+2n(pi)}{3}
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.

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

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

Last edited:
thanks - i) makes sense.

so ii) -> 0 ?

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

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

Mattofix said:
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, xn, and x'n), both of which grow without bounds, and sin(xn) = 0, sin(x'n) = 1, for all n.

We can choose xn = 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? :)

Last edited:
yep - thanks

## What is the purpose of finding the limits of a function?

The limit of a function is the value that the function approaches as the input approaches a certain value. It helps us understand the behavior of the function and its values near a specific point.

## How do I find the limit of a function algebraically?

To find the limit of a function algebraically, you can use techniques such as direct substitution, factoring, rationalization, and L'Hôpital's rule. These methods involve manipulating the function to make the limit easier to evaluate.

## What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the behavior of the function as the input approaches the limit from one side (left or right). A two-sided limit takes into account the behavior of the function from both sides of the limit.

## Can the limit of a function exist even if the function is not defined at that point?

Yes, it is possible for the limit of a function to exist at a point where the function is not defined. This is because the limit only considers the behavior of the function near that point, not necessarily the value at that point.

## Why is it important to consider the limit of a function when there is a discontinuity?

When a function has a discontinuity, it means that there is a break or gap in the graph of the function. By finding the limit at the point of discontinuity, we can better understand the behavior of the function and determine if the function is continuous or not at that point.

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