i got 3.434 which is wreong i got 1 which was wrong?? very confuised
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Not sure about the first one, but for the second one, you need to use L'hopital's rule, and you should get 6/5.
Originally Posted by kyleu03 i got 3.434 which is wreong i got 1 which was wrong?? very confuised $\displaystyle \lim_{x \to 0} \frac{\tan(6x)}{\tan(2x)} $ use L'Hopital ... $\displaystyle \lim_{x \to 0} \frac{6\sec^2(6x)}{2\sec^2(2x)} = \frac{6}{2} = 3 $ hint for #2 ... $\displaystyle x^n-1 = (x-1)(x^{n-1} + x^{n-2} + ... + x^2 + x + 1)$
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