lim x-->0 ##\frac{x tan2x -2xtanx} {(1-cos2x)^2}##

[tellmesomething]

Homework Statement

Title

Relevant Equations

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I simplified this function to
$\frac{1}{2} (\frac{x tan^3(x)} {(sin²x)²(1-tan²x)}$
Now further can I not write $1-tan²x$ as $\frac{cos2x} {sin²x}$ ?

If I do that I get $\frac{1}{2} (\frac{x tan^3(x)} {sin²x cos2x}$

On graphing this on desmos I get two different graphs for these functions

My first simplification matches with the original function but the second one doesnt and hence doesnt fetch me the correct liming value. whys this?

 

[anuttarasammyak]

Why don’t you apply L'hopital's Rule ? Or Taylor’s expansion. X^4 order seems to mater.

 

[tellmesomething]

Why don’t you apply L'hopital's Rule ? Or Taylor’s expansion. X^4 order seems to mater.

Yes but I dont get why my simplification yields a different function? Is it because the domain has changed of the function?(if it has at all)

 

[vela]

My first simplification matches with the original function but the second one doesnt and hence doesnt fetch me the correct liming value. whys this?

Obviously, you made a mistake somewhere. I don't know how you expect us to find the mistake if you don't show us your work.

 

[FactChecker]

Now further can I not write $1-tan²x$ as $\frac{cos2x} {sin²x}$ ?

Evaluate these two at x=0. They are not equal.

 

[anuttarasammyak]

I got

Spoiler

$$\frac{1}{2}\frac{x}{\sin x}\frac{1}{(2\cos^2 x-1)\cos x}$$

I hope it will help your check.