I would appreciate if someone could verify to me my answers.
limx→0x−sin(x)tan(x)−x
I used L'Hopital's rule twice and got answer 1/2.
limx→0+1sin(4x)−14x
also I used L'Hopital's rule twice and got 0.
limx→2−(x2−4)ln(2−x)
I used L'Hopital's rule once and got 0.
Thanks.
Answer
Since the answer to the first problem has been given by GTX OC, let us focus on the second problem.
(x2−4)log(2−x)=(x−2)(x+2)log(2−x)=−(x+2)(2−x)log(2−x)
I suppose you know that xlog(x) goes to 0 when x goes to zero. Then, ... Are you able to continue with this ?
No comments:
Post a Comment