Here's the question you clicked on:
kvdop
Find the lim of tan(2x)/(3x+sinx) as x approaches 0
I don't see any quick way without l'Hopital.
l'Hopitals has taken me back to the start
If l'Hopital doesn't work the first time but gives you the condition for it again, you can repeat it. You might have better luck the second time around.
One way is to use l'Hopital, but a faster way is to use the linear approximations for sine and tangent. The answer is 1/2.
THE DIFFERENTIAL IS SEC^2(X)/3+COS(x).numerator is 1 and denominator is 4 answer is 1/4 how did you get 1/2??????
pasta! skywalker94 is right! Using l'Hopital, we'll get 1/2. \[\lim_{x \rightarrow 0}(\tan2x/(3x+sinx))=\lim_{x \rightarrow 0}[2(1+\tan^{2}2x)/(3+cosx)]=1/2\]
use expansion method it is the fastest.... tanx = x + x^3/3............sinx = x - x^3/3!
and from that the ans is 1/2 Math Physics and skywalker ur right..!!!!