Calculating a limit with tangent as denominator

In summary, the problem is to find the limit of (x^3-2x^2+x)/tanx as x approaches 0. The attempt at a solution involves factoring out an x from the numerator and rearranging the expression to apply the property that the limit of a product is equal to the product of the limits. The final answer is 1, but the process of getting there is not known.
  • #1
Tebow15
10
0

Homework Statement


lim x^3-2x^2+x/tanx
x->0


The Attempt at a Solution



All i know is that tan is going to break up into sinx/cosx so the equation will look like this

lim x^3-2x^2+x/(sinx/cosx)
x->0

I haven't worked with cubic or quadratic functions yet so I don't know where to begin with this.
 
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  • #2
Tebow15 said:

Homework Statement


lim x^3-2x^2+x/tanx
x->0

Remember to throw in brackets where necessary, because the way it looks right now is

[tex]\lim_{x\to 0}\left(x^3-2x^2+\frac{x}{\tan(x)}\right)[/tex]

So you need to put brackets around the cubic to avoid this confusion :wink:
Tebow15 said:

The Attempt at a Solution



All i know is that tan is going to break up into sinx/cosx so the equation will look like this

lim x^3-2x^2+x/(sinx/cosx)
x->0

I haven't worked with cubic or quadratic functions yet so I don't know where to begin with this.

If your problem is of the form [tex]\lim_{x\to a}\left(f(x)\cdot g(x)\right)[/tex] then this is equivalent to [tex]\lim_{x\to a}f(x)\cdot\lim_{x\to a} g(x)[/tex]

So try factoring out an x from the numerator, and see if you can apply this idea correctly.
 
  • #3
I may have confused you with the way I wrote it, it should look like this. (I'll use brackets to make it clearer)

lim (x^3-2x^2+x)/(sinx/cosx)
x->0
 
  • #4
The answer eventually = 1 but I don't know how to get there
 
  • #5
Factor out x as Mentallic suggested and rearange the expression as

[tex]\left(\frac{x}{\sin(x)}\right)(x^2-2x+1)\cos(x)[/tex]

ehild
 

What is a limit with tangent as denominator?

A limit with tangent as denominator is a mathematical concept used to describe the behavior of a function as the input values approach a specific point or value. In this case, the function's denominator (bottom) contains a tangent function.

How do you calculate a limit with tangent as denominator?

To calculate a limit with tangent as denominator, you can use the fact that the limit of a quotient is equal to the quotient of the limits. This means that you can take the limit of the numerator and the limit of the denominator separately and then divide the two results.

Can the limit of a function with tangent as denominator be undefined?

Yes, the limit of a function with tangent as denominator can be undefined if the limit of the numerator and the limit of the denominator are both equal to zero.

Are there any special cases when calculating a limit with tangent as denominator?

Yes, there are some special cases when calculating a limit with tangent as denominator. For example, if the limit of the numerator and the limit of the denominator are both equal to zero, you may need to use L'Hopital's rule to solve the limit. Additionally, if the limit of the denominator is equal to infinity, the limit of the function may also be undefined.

How is a limit with tangent as denominator used in real-life applications?

A limit with tangent as denominator is used in many real-life applications, such as in physics, engineering, and economics. For example, it can be used to describe the rate of change of a variable, the speed of an object, or the growth of a population. It is also used in the study of calculus, which is essential in many fields of science and technology.

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