Question

Limit x-0 [sin(α+β)x + sin(α-β)x + 2sinαx]/cos2βx-cos2αx *x

Answer 1

Answer:

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

Answer 2

2 is right ans bro I try my best
hope it helps
you