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i know if you just plug in the value you get 0/0 so you use the rule
so i use the rule and get |dw:1361572627877:dw|
how do i get from here as you cant use the trigonometric substitution either
the derivative of tan is sec^2 x lol
so thats wrong
|dw:1361572818037:dw| then differentiate from here
the limit goes to -infinity
Change your 1 -sec^2x into -tan^2x and see if that helps you simplify your ratio.
thats not how it works youcan never change 1 - sec^2 into tan^2x
you can change it into -tan^2x though, not tan^2x
ok im gonna try walters method
But that only makes it worse. Nevermind. Hang on.
you dont understand tan^2 x + 1 = sec^2x try to change it to what you just did it never works
Move the 1 over, and then multiply both sides by a -1.
i thought about your method walter but its x not 1 - tanx
Without using L'hopital |dw:1361573289043:dw|
so the solution is 1/0 but thats not gonna work either
or I could have misunderstood your question http://www.wolframalpha.com/input/?i=lim+x-%3E0+sin%28x%29%2F%28x-tan%28x%29%29
If you use L'hopital's rule twice, you end up with the limit as x approaches negative infinity of cos^3x/(-2)
no that is the question experimentX
oops, it's approaching zero, so the answer is -1/2
You can't use L'hopital rule twice, one you use L'hopital rule, ... the top is cos(x) ... which is 1 and the bottom part is still 0.
Ah hell, then it's no longer of indeterminate form.
the best way see through limits is to expand using Taylor series. Try expanding it ... the answer is pretty obvious. Anyway gotta sleep ... best of luck.
i think i got it now ty a lot
I need to go refresh my memory. Good luck to you.