# Finding the limit using a trig identity

• ver_mathstats
In summary, the limit as x approaches 0 of x2/(sin2x(9x)) is 1/81, but if you multiply by x/x, you get this limit: ##\lim_{x \to 0} \frac x {9x^2} \cdot \frac {x^2}{\sin^2(x)}##, which still doesn't exist.
ver_mathstats
Homework Statement
Find the limit as x approaches 0 x^2/(sin(^2)x(9x))
Relevant Equations
limit as x approaches 0 x^2/(sin(^2)x(9x))
Find the limit as x approaches 0 of x2/(sin2x(9x))

I thought I could break it up into:
limit as x approaches 0 ((x)(x))/((sinx)(sinx)(9x)).

So that I could get:
limx→0x/sinx ⋅ limx→0x/sinx ⋅ limx→01/9x.

I would then get 1 ⋅ 1 ⋅ 1/0. Meaning it would not exist.

However the solution is 1/81 in the textbook, would this mean I would have to multiply the numerator and denominator, specifically 1/9x by x/x to get x/9x2. If so, why would I have to do this? If this is wrong how would I approach this then?

Thank you.

Find the correct problem statement, perhaps something like $$\lim_{x\downarrow 0} {x^2\over \sin^2(9x)}$$ and learn to present it

ver_mathstats said:
Homework Statement: Find the limit as x approaches 0 x^2/(sin(^2)x(9x))
Homework Equations: limit as x approaches 0 x^2/(sin(^2)x(9x))

Find the limit as x approaches 0 of x2/(sin2x(9x))

I thought I could break it up into:
limit as x approaches 0 ((x)(x))/((sinx)(sinx)(9x)).

So that I could get:
limx→0x/sinx ⋅ limx→0x/sinx ⋅ limx→01/9x.

I would then get 1 ⋅ 1 ⋅ 1/0. Meaning it would not exist.

However the solution is 1/81 in the textbook, would this mean I would have to multiply the numerator and denominator, specifically 1/9x by x/x to get x/9x2. If so, why would I have to do this? If this is wrong how would I approach this then?

Thank you.
If you multiply by x/x, you get this limit: ##\lim_{x \to 0} \frac x {9x^2} \cdot \frac {x^2}{\sin^2(x)}##, which still doesn't exist.

Can you upload an image of the problem as shown in your textbook? I suspect that there is either another factor of 9 that you don't show, or that there is a typo in your book.

BTW, this -- limit as x approaches 0 x^2/(sin(^2)x(9x)) -- is really hard to read. It can be written much more clearly using LaTeX. We have an informative tutorial at the link shown at the bottom of the text entry pane.

Mark44 said:
If you multiply by x/x, you get this limit: ##\lim_{x \to 0} \frac x {9x^2} \cdot \frac {x^2}{\sin^2(x)}##, which still doesn't exist.

Can you upload an image of the problem as shown in your textbook? I suspect that there is either another factor of 9 that you don't show, or that there is a typo in your book.

BTW, this -- limit as x approaches 0 x^2/(sin(^2)x(9x)) -- is really hard to read. It can be written much more clearly using LaTeX. We have an informative tutorial at the link shown at the bottom of the text entry pane.
Okay I'm not sure if I inserted the picture correctly, hopefully it worked, and yes thank you, I didn't really know how to write it properly but I am going to learn now using LaTeX. I realized I had made an error in writing it out and that is why I kept struggling with it, I am very sorry. I could break it up into x/sin(9x) ⋅ x/sin(9x) and multiply each side by 9/9 to achieve 9/9 ⋅ x/sin(9x) and then I factor out the 1/9 so I am left with 1/9 limx→0 9x/sin(9x) which is 1/9 ⋅ 1. However, I do this two times and I am then left with 1/81. Thank you for the reply.

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BvU said:
Find the correct problem statement, perhaps something like $$\lim_{x\downarrow 0} {x^2\over \sin^2(9x)}$$ and learn to present it
Thank you for the reply, I had realized I made an error while writing out the question.

ver_mathstats said:
Okay I'm not sure if I inserted the picture correctly, hopefully it worked,
Yes, it worked.
ver_mathstats said:
and yes thank you, I didn't really know how to write it properly but I am going to learn now using LaTeX.
It's really not all that complicated. Here is the limit in the image you posted, using LaTeX:
$$\lim_{x \to 0}\frac{x^2}{\sin^2(9x)}$$

Here is the unrendered script I used:
$$\lim_{x \to 0}\frac{x^2}{\sin^2(9x)}$$

The trick to evaluating this limit is to multiply by 1 in the form of 81/81.

$$\lim_{x \to 0}\frac{(9x)^2}{\sin^2(9x)} \cdot \frac 1 {81}$$

You can break up the above into two limits, with the first limit being 1, and the second being 1/81.

## 1. What is a trig identity?

A trig identity is a mathematical equation that expresses a relationship between different trigonometric functions. Examples of trig identities include the Pythagorean identity, the double angle identities, and the product-to-sum identities.

## 2. How do you use a trig identity to find the limit of a function?

To find the limit of a function using a trig identity, you first need to rewrite the function using the trig identity. Then, you can use algebraic manipulation and trigonometric properties to simplify the expression and evaluate the limit. This method is especially useful for evaluating limits involving trigonometric functions that approach indeterminate forms such as 0/0 or ∞/∞.

## 3. Can a trig identity be used to find the limit of any function?

No, a trig identity can only be used to find the limit of a function that contains trigonometric functions. It cannot be used for other types of functions such as polynomial or exponential functions.

## 4. What are some common trig identities used for finding limits?

Some common trig identities used for finding limits include the Pythagorean identity, which relates sine and cosine, and the double angle identities, which can be used to simplify expressions involving trigonometric functions. Other useful identities include the sum and difference identities, the half angle identities, and the product-to-sum identities.

## 5. Are there other methods for finding limits besides using trig identities?

Yes, there are other methods for finding limits, such as using L'Hôpital's rule, the squeeze theorem, or direct substitution. However, using trig identities can be a helpful approach for evaluating limits of trigonometric functions, especially when the other methods do not apply or are more complicated.

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