View Full Version : Limit with trig function
theCandyman
Dec6-04, 02:29 PM
I am completly lost on where to start, any help would be appreciated:
lim (x->0) tan^2(2x) / 3x^2
I do not have any work so far, but I'm trying to find a way to get rid of the x^2 in the denominator, which so far has been fruitless.
Try replacing tan(2x) with sin(2x)/cos(2x). You probably know something about the limit of sin(x)/x as x->0.
dextercioby
Dec6-04, 02:50 PM
I am completly lost on where to start, any help would be appreciated:
lim (x->0) tan^2(2x) / 3x^2
I do not have any work so far, but I'm trying to find a way to get rid of the x^2 in the denominator, which so far has been fruitless.
This limit works fine without the aid of Mr.L'Ho^pital.
Are u 100% convinced that:
\lim_{2x \rightarrow 0} \frac{\tan{2x}}{2x} =1
????????????
If so,u'll be able to apply the formula above properly and get the result.
One more hint:
\lim_{x \rightarrow 0} \frac{\tan^{2} 2x}{3x^2} = \frac{4}{3}
(\lim_{2x \rightarrow 0} \frac{\tan{2x}}{2x})^{2}
I hope you can take it from there.
theCandyman
Dec7-04, 03:29 AM
Thank you both, I have my answer now.
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