# Lim y->0: Solving (sin 3y * cot 5y) / (y * cot 4y)

• Oneiromancy
I meant y instead of x.\lim_{y\rightarrow 0}\frac{\sin{3y}}{y}\cdot\frac{\sin{4y}}{\cos{4y}}\cdot\frac{\cos{5y}}{\sin{5y}}In summary, the limit of (sin 3y * cot 5y)/(y * cot 4y) as y approaches 0 can be simplified to (sin 3y * sin 4y * cos 5y)/(y * sin 5y * cos 4y). However, the left term is still indeterminate as y approaches 0.
Oneiromancy
lim y->0 (sin 3y * cot 5y)/(y * cot 4y)

Tricky problem to me. I understand what to do if the problem was something like sin 5x / 4x.

Last edited:
Well first rewrite cot in terms of sin and cosine, and look to come up with something sinx/x when x-->0 to some parts of it at least.

$$\lim_{y\rightarrow 0}\frac{\sin{3y}\cot{5y}}{y\cot{4y}}$$

rocophysics said:
$$\lim_{y\rightarrow 0}\frac{\sin{3y}\cot{5y}}{y\cot{4y}}$$

Correct. Sorry I'm not good at latex.

$$\lim_{y\rightarrow 0}\frac{\sin{3y}}{y}\cdot\frac{\sin{4y}}{\cos{4y}}\cdot\frac{\cos{5y}}{\sin{5y}}$$

How can you manipulate your limit? The left term is very easy.

Last edited:
I know what to do now.

rocophysics said:
$$\lim_{y\rightarrow 0}\frac{\sin{3x}}{y}\cdot\frac{\sin{4y}}{\cos{4y}}\cdot\frac{\cos{5y}}{\sin{5y}}$$

How can you manipulate your limit? The left term is very easy.

It would be easier if it were sin(3y)/y rather than sin(3x)/y !

HallsofIvy said:
It would be easier if it were sin(3y)/y rather than sin(3x)/y !
It was a typo!

## 1. What is the limit of (sin 3y * cot 5y) / (y * cot 4y) as y approaches 0?

The limit of (sin 3y * cot 5y) / (y * cot 4y) as y approaches 0 is indeterminate. This means that it does not have a single, finite value. Instead, it can approach different values depending on how the expression is simplified.

## 2. How can I simplify the expression to find the limit?

To simplify the expression, you can use trigonometric identities and algebraic manipulation. For example, you can rewrite cot 5y as cos 5y / sin 5y and cot 4y as cos 4y / sin 4y. Then, you can use the fact that sin 3y / y approaches 1 as y approaches 0 and cos 3y / y approaches 0 as y approaches 0. This will help you simplify the expression and find the limit.

## 3. Can I use L'Hopital's rule to find the limit?

Yes, you can use L'Hopital's rule to find the limit. This rule states that if you have an indeterminate form, such as 0/0 or ∞/∞, you can take the derivative of the numerator and denominator separately and then evaluate the limit again. This can help simplify the expression and find the limit.

## 4. What is the difference between a removable and non-removable discontinuity?

A removable discontinuity occurs when the limit of a function exists but is not equal to the value of the function at that point. This can be fixed by redefining the function at that point. On the other hand, a non-removable discontinuity occurs when the limit of a function does not exist at a certain point. This can happen when there is a vertical asymptote or a jump in the graph of the function at that point.

## 5. How can I graph the function to better understand the limit?

To graph the function and better understand the limit, you can use a graphing calculator or software. You can also use the algebraic simplifications and trigonometric identities mentioned earlier to rewrite the expression in a way that is easier to graph. Additionally, you can plot points on the graph to see how the function behaves near the point in question and estimate the limit.

• Calculus and Beyond Homework Help
Replies
7
Views
835
• Calculus and Beyond Homework Help
Replies
16
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
27
Views
3K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Precalculus Mathematics Homework Help
Replies
25
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
910
• Calculus and Beyond Homework Help
Replies
4
Views
1K