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lim [(x^2+3x-1)tanx]/(x^2+2x) . x->0
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= (1 + 3/x - 1/x^2)tanx/(1 + 2/x) = (1 + 0 - 0)*0/(1 + 0) = 0
Answer is supposed to be -0,5
note that \[\lim_{x\to 0}\frac{\tan(x)}{x}=1\]
see the attachment ^
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First step I multiplied and divided everything by "x" so that I get the limit which Zarkon here said ;)
Thanks!
You're welcome :)
damn. where did i go wrong? i can't see it
You have a division by zero there, check the 1+2/x
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oh wait, i'm a complete idiot. i evaluated the limit of 2/x to be 0...
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