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evaluate the limit lim cos(7x) -1 / x^2 x->0
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so far i've done -7 sin 7x -1 / 2x
using the double angle formula, we have \(cos 2\theta=1-2\sin^2\theta\)
need to use l hospital's rule to evaluate but i'm not sure how to work it out
no need for that.. just use the double angle formula
\[ L=\lim_{x\to0}\frac{1-2\sin^2\left(7x\over2\right)-1}{x^2} \]
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follow?
no I ended up with -7cos(7x) / 2 = 7/2
our whole point was to get the sine
use the equation editor below to enter your steps.
the answer is 49/2
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i mean -49/2
we are trying to create the form \[ \lim_{x\to0}\frac{\sin x}{x}=1 \]
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