Stuck on trig limit: lim[x->0](sin^2(2x))/(tan(x)*tan^2(3x)) = 0/0 = infty...?

[Beest]

Hey guys! First post here, and I want to clear this up: my education is in Dutch, so I'm not very familiar with all the English mathematical vernacular. I just googled everything so I hope I don't sound like a moron.

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\(\displaystyle \displaystyle \lim_{x \rightarrow 0}\, \dfrac{\sin^2(2x)}{\tan(x)\, \cdot\, \tan^2(3x)}\, =\, \dfrac{0}{0}\)

\(\displaystyle \displaystyle\, =\, \dfrac{4}{9}\, \left(\lim_{x \rightarrow 0}\, \dfrac{\sin(2x)}{2x}\right)^2\, \cdot\, \left(\lim_{x \rightarrow 0}\, \dfrac{3x}{\tan(3x)}\right)^2\, \cdot\, \left(\lim_{x \rightarrow 0}\, \dfrac{1}{\tan(x)}\right)\)

\(\displaystyle \displaystyle\, =\, \dfrac{4}{9}\, \cdot\, 1\, \cdot\, 1\, \cdot\, \dfrac{1}{0}\, =\, \infty\)

\(\displaystyle \displaystyle \mbox{LL }\, =\, \lim_{x \rightarrow 0^-}\, \dfrac{\sin^2(2x)}{\tan(x)\, \cdot\, \tan^2(3x)}\, =\, -\infty\)

\(\displaystyle \displaystyle \mbox{RL }\, =\, \lim_{x \rightarrow 0^+}\, \dfrac{\sin^2(2x)}{\tan(x)\, \cdot\, \tan^2(3x)}\, =\, +\infty\)

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The exercise that I attached is a solution by my prof but it's got me completely stumped, I can't wrap my head around that first conversion.
How did the 2x in the first denominator come to be? And the 3x in the second numerator? I suppose it has something to do with the 4/9 at the start, but I simply can't connect the dots here.

Can someone please point me in the right direction by inserting a step?

Many thanks!
-Phil

 

[stapel]

Hey guys! First post here, and I want to clear this up: my education is in Dutch, so I'm not very familiar with all the English mathematical vernacular. I just googled everything so I hope I don't sound like a moron.

You're doing great!

\(\displaystyle \displaystyle \lim_{x \rightarrow 0}\, \dfrac{\sin^2(2x)}{\tan(x)\, \cdot\, \tan^2(3x)}\, =\, \dfrac{0}{0}\)

\(\displaystyle \displaystyle\, =\, \dfrac{4}{9}\, \left(\lim_{x \rightarrow 0}\, \dfrac{\sin(2x)}{2x}\right)^2\, \cdot\, \left(\lim_{x \rightarrow 0}\, \dfrac{3x}{\tan(3x)}\right)^2\, \cdot\, \left(\lim_{x \rightarrow 0}\, \dfrac{1}{\tan(x)}\right)\)

\(\displaystyle \displaystyle\, =\, \dfrac{4}{9}\, \cdot\, 1\, \cdot\, 1\, \cdot\, \dfrac{1}{0}\, =\, \infty\)

\(\displaystyle \displaystyle \mbox{LL }\, =\, \lim_{x \rightarrow 0^-}\, \dfrac{\sin^2(2x)}{\tan(x)\, \cdot\, \tan^2(3x)}\, =\, -\infty\)

\(\displaystyle \displaystyle \mbox{RL }\, =\, \lim_{x \rightarrow 0^+}\, \dfrac{\sin^2(2x)}{\tan(x)\, \cdot\, \tan^2(3x)}\, =\, +\infty\)

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The exercise that I attached is a solution by my prof but it's got me completely stumped, I can't wrap my head around that first conversion.

That's because your professor did some "magic", using limits that you've probably forgotten; namely:

. . . . .\(\displaystyle \displaystyle \lim_{x \rightarrow 0}\, \dfrac{\sin(x)}{x}\, =\, \lim_{x \rightarrow 0}\, \dfrac{x}{\sin(x)}\, =\, 1\)

. . . . .\(\displaystyle \displaystyle \lim_{x \rightarrow 0}\, \dfrac{\tan(x)}{x}\, =\, \lim_{x \rightarrow 0}\, \dfrac{x}{\tan(x)}\, =\, 1\)

Having this in the back of his head, he split the fraction apart into:

. . . . .\(\displaystyle \left(\dfrac{\sin(2x)}{1}\right)^2\, \cdot\, \left(\dfrac{1}{\tan(3x)}\right)^2\, \cdot\, \left(\dfrac{1}{\tan(x)}\right)\)

The first and second fractions aren't yet set up for application of the known limits, so he put them in, making sure to multiply fractions by useful forms of "1", so as not to change any values:

. . . . .\(\displaystyle \left(\dfrac{\sin(2x)}{2x}\, \cdot\, \dfrac{2x}{1}\right)^2\, \cdot\, \left(\dfrac{3x}{\tan(3x)}\, \cdot\, \dfrac{1}{3x}\right)^2\, \cdot\, \left(\dfrac{1}{\tan(x)}\right)\)

Then he took the parts that were unnecessary for application of the known limits:

. . . . .\(\displaystyle \left(\dfrac{2x}{1}\, \cdot\, \dfrac{1}{3x}\right)^2\, \cdot\, \left(\dfrac{\sin(2x)}{2x}\right)^2\, \cdot\, \left(\dfrac{3x}{\tan(3x)}\right)^2\, \cdot\, \left(\dfrac{1}{\tan(x)}\right)\)

Then he simplified the stuff in the first parenthetical above. Then he applied (loosely, I think?) the product property of limits. (here)

Hope that helps!

 

[Beest]

Thanks so much! That helped a lot. What an Aha-experience!

P.S: Did you just assume my prof's gender?! She won't like that! hahaha

 

[mmm4444bot]

Did you just assume my prof's gender?

I'll answer, "nah", for stapel, despite knowing that she can well answer this herself.

The choice of wording is similar to yours, when you "assumed" (not really) that we're all males here.

Hey guys!

 

[Beest]

Way to overthink a joke mate.

 

[mmm4444bot]

Way to overthink a joke mate.

Oh, I understood your wit, but you seem to have missed mine!

I simply turned your joke around, so it's on you. I hope that I did not upset your day; keep up the good work! :cool: