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

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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
Screenshot_2024-07-11-19-08-34-995_com.desmos.calculator.jpg

My first simplification matches with the original function but the second one doesnt and hence doesnt fetch me the correct liming value. whys this?
 
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Why don’t you apply L'hopital's Rule ? Or Taylor’s expansion. X^4 order seems to mater.
 
anuttarasammyak said:
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)
 
tellmesomething said:
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.
 
tellmesomething said:
Now further can I not write ##1-tan²x## as ##\frac{cos2x} {sin²x}## ?
Evaluate these two at x=0. They are not equal.
 
I got
\frac{1}{2}\frac{x}{\sin x}\frac{1}{(2\cos^2 x-1)\cos x}
I hope it will help your check.
 
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