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

  • Thread starter Oneiromancy
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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.
  • #1
Oneiromancy
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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.
 
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  • #2
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.
 
  • #3
[tex]\lim_{y\rightarrow 0}\frac{\sin{3y}\cot{5y}}{y\cot{4y}}[/tex]
 
  • #4
rocophysics said:
[tex]\lim_{y\rightarrow 0}\frac{\sin{3y}\cot{5y}}{y\cot{4y}}[/tex]

Correct. Sorry I'm not good at latex.
 
  • #5
[tex]\lim_{y\rightarrow 0}\frac{\sin{3y}}{y}\cdot\frac{\sin{4y}}{\cos{4y}}\cdot\frac{\cos{5y}}{\sin{5y}}[/tex]

How can you manipulate your limit? The left term is very easy.
 
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  • #6
I know what to do now.
 
  • #7
rocophysics said:
[tex]\lim_{y\rightarrow 0}\frac{\sin{3x}}{y}\cdot\frac{\sin{4y}}{\cos{4y}}\cdot\frac{\cos{5y}}{\sin{5y}}[/tex]

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 !
 
  • #8
HallsofIvy said:
It would be easier if it were sin(3y)/y rather than sin(3x)/y !
It was a typo!
 

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

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.

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.

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.

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.

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.

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