# Limits Homework Help: Questions on \lim_{x\rightarrow 0}

• duki
In summary, the conversation discusses two problems related to limits, with the first problem involving the use of the sin formula to find the correct answer of 4/5. The second problem involves rationalizing the expression to simplify it and find the limit to be 1. The conversation also briefly touches on the use of rationalizing in both the top and bottom of an expression.
duki

## Homework Statement

Have a couple of questions actually.

Problem #1 $$\lim_{x\rightarrow 0}{\left(\frac{4x}{\sin 5x}\right)}$$

## Homework Equations

I was told in class to do $$\lim_{x\rightarrow 0}{\left(\frac{4x*5}{5\sin 5x}\right)}$$

## The Attempt at a Solution

so after that I was told that $${\left(\frac{x}{\sin x}\right)} = 1$$
but my book says $${\left(\frac{x}{\sin x}\right)} = 0$$

The final answer we were given was $$4/5$$

Could someone explain this to me? I know how he got 4/5, but which sin formula is correct?------------------

Problem #2 $$\lim_{x\rightarrow 0}{\left(\frac{1-\cos x}{x^2}\right)}$$

I can get to this step:

$$\lim_{x\rightarrow 0}{\left(\frac{\sin^2 x}{\left(1+\cos x\right)x^2}\right)}$$

but what's next?

Thanks guys! I have a test tomorrow, so I'm just trying to get some things straight

as x-->0 sinx/x=1 so that as x-->0 x/sinx =1

For the second one put

$$\frac{sin^2x}{(1+cosx)x^2}=(\frac{sinx}{x})^2 \frac{1}{1+cosx}$$

Last edited:
x / sin x can become 1 / (sin x / x) which equals 1/1 = 1.

well you almost got it, you did the most part, here

$$\lim_{x\rightarrow 0}{\left(\frac{4x*5}{5\sin 5x}\right)}=\frac{4}{5}\lim_{x\rightarrow 0}{\left(\frac{5x}{\sin 5x}\right)}=\frac{4}{5}\lim_{x\rightarrow 0}{\left(\frac{1}{\frac{\{sin 5x}{\ 5x}\right)}}=$$ no remember that

lim(x-->0)sinB/B=1, where B in our case is 5x

so $$\lim_{x\rightarrow 0}\frac{\sin x}{x}} = \frac{x}{\sin x}} = 1$$?

well yeah, and don't forget the limit sign, otherwise it is not true,lol.

you're fast.
Great, that is all I needed for that one I suppose. Thanks.

For the second one, are you just factoring out the ^2 so you can get 1?

In a basic sense, yes

groovy. This one's solved. thanks a bunch guys :D:D:D

Another quick question.

I have $$\lim_{x\rightarrow -1}{\left(\frac{\sqrt {x^2 + 8} + 3}{x+1}\right)}$$

I know you multiply by $$\sqrt {x^2 + 8} - 3$$ but what if the +3 wasn't there? Or what if it were on the bottom?

duki said:
Another quick question.

I have $$\lim_{x\rightarrow -1}{\left(\frac{\sqrt {x^2 + 8} + 3}{x+1}\right)}$$

I know you multiply by $$\sqrt {x^2 + 8} - 3$$ but what if the +3 wasn't there? Or what if it were on the bottom?

If it was on the buttom then that limit would be 1, and if it wasnt there then that limit would not exist. Because the two sided limits would not match.

Well just for demonstration purposes? Why do we use - 3 but not - 8 ?

And if it were on the bottom, would it be the same process (assuming there was a limit)

The reason why we multiply the bottom and the top by $$\sqrt {x^2 + 8} - 3$$
is because we want to form a difference of squares in the top, that is somethign like this a^2-b^2=(a-b)(a+b),so we can cancele some terms out and thus end up with something that can cancle out with the bottom, this way we manage to facilitate the process.
Which, if it were in the bottom, the whole expression $$\sqrt {x^2 + 8} + 3$$??

Usually we rationalise either the bottom or the top when we have some simple intermediate forms like 0/0 or infinty/infinity, then sometimes by just rationalizing the bottom or the top we get rid of these intermediate forms, but not always.

Can you be more specific, what are u really asking?

## 1. What is a limit in calculus?

A limit in calculus represents the value that a function approaches as the input approaches a certain value. It is denoted by the symbol \lim_{x\rightarrow a}, where x is the input and a is the value the input is approaching.

## 2. How do I evaluate a limit?

To evaluate a limit, you can try plugging in the value the input is approaching into the function and see if it approaches a specific value. If it does, then that value is the limit. If it does not, you may need to use algebraic manipulations or special limit rules to evaluate the limit.

## 3. What are the different types of limits?

There are three types of limits: one-sided limits, two-sided limits, and infinite limits. One-sided limits only consider the value of the function as the input approaches from one side, while two-sided limits consider both sides. Infinite limits occur when the value of the function approaches positive or negative infinity.

## 4. What is the difference between a limit and a derivative?

A limit and a derivative are both concepts in calculus, but they have different purposes. A limit represents the value a function approaches, while a derivative represents the rate of change of a function at a specific point.

## 5. How can I use limits to solve real-life problems?

Limits are used in many real-life applications, such as calculating the maximum height of a projectile or determining the maximum speed of a moving object. They can also be used to model population growth and predict future values based on current trends.

• Calculus and Beyond Homework Help
Replies
17
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
472
• Calculus and Beyond Homework Help
Replies
10
Views
975
• Calculus and Beyond Homework Help
Replies
8
Views
917
• Calculus and Beyond Homework Help
Replies
10
Views
977
• Calculus and Beyond Homework Help
Replies
14
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
496
• Calculus and Beyond Homework Help
Replies
12
Views
1K
• Calculus and Beyond Homework Help
Replies
4
Views
1K