Find the limit. Use l'Hospital's Rule if appropriate.
lim (x-sinx)/(x-tanx)
x->0
I've tried this problem a couple times and I just can't quite get it, I know L'Hospital's Rule, i just need to know the procedure to the solution. Trig has never been a strong part of mine so that might be part of the problem. Thanks in advance.
lim (x-sinx)/(x-tanx)
x->0
I've tried this problem a couple times and I just can't quite get it, I know L'Hospital's Rule, i just need to know the procedure to the solution. Trig has never been a strong part of mine so that might be part of the problem. Thanks in advance.
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lim x->0 (x - sin(x))/(x - tan(x))
The tan(x) = sin(x)/cos(x) rule won't really help us here much because of the x we have in both the numerator and denominator. Let's just use L'Hopitals rule directly by taking the derivative of the numerator and the denominator:
lim x->0 (1 - cos(x))/(1 - sec²(x)) = lim x->0 (cos²(x))(cos(x) - 1)/(1 + cos(x))(1 - cos(x))
= lim x->0 -cos²(x)/(1 + cos(x)) = -1/(1 + 1) = -1/2
The tan(x) = sin(x)/cos(x) rule won't really help us here much because of the x we have in both the numerator and denominator. Let's just use L'Hopitals rule directly by taking the derivative of the numerator and the denominator:
lim x->0 (1 - cos(x))/(1 - sec²(x)) = lim x->0 (cos²(x))(cos(x) - 1)/(1 + cos(x))(1 - cos(x))
= lim x->0 -cos²(x)/(1 + cos(x)) = -1/(1 + 1) = -1/2
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tan (x) = sin(x) / cos(x) if that helps any.