Finding the limit using a trig identity

[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:
lim_x→0x/sinx ⋅ lim_x→0x/sinx ⋅ lim_x→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.

 

[BvU]

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

 

[Mark44]

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:
lim_x→0x/sinx ⋅ lim_x→0x/sinx ⋅ lim_x→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.

 

[ver_mathstats]

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 lim_x→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.

 

[ver_mathstats]

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.

 

[Mark44]

Okay I'm not sure if I inserted the picture correctly, hopefully it worked,

Yes, 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.

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.