Limit x-0 [sin(α+β)x + sin(α-β)x + 2sinαx]/cos2βx-cos2αx *x
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Step-by-step explanation:
You can rewrite your function as
βx3αx−sinxsin(βx)βx
If α≠1, this also becomes
βx2α−sinxxsin(βx)βx
Note that the second factor has limit 1 and the first factor has limit 0 (the numerator has limit 0; the denominator has limit α−1≠0).
Thus if the limit has to be 1, we need α=1.
Now it should be well known that
limx→0x−sinxx3=16
(apply l'Hôpital or a Taylor expansion), so we have
limx→0βx−sinxx3sin(βx)βx=6β
Now it's just an easy computation
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