Question

find the "a" value: lim (1-cosax)/(xsin3x)=1/6
x-->0

Answer 1

limit what \(\mbox{(-_-'')}\)

Answer 2

x tends to what

Answer 3

sorry it is "0"

Answer 4

Is L'hopital rule taught?

Answer 5

I don't know, just need help

Answer 6

-_-

Answer 7

Apply differentiation to the numerator and the denominator respectively and together

Answer 8

I need answer, please

Answer 9

http://www.wolframalpha.com/input/?i=lim+x-%3E0+%281-cos+ax%29%2F%28xsin3x%29%3D1%2F6%2C+find+a

Answer 10

how did you find it?

Answer 11

sorry I don't know :/

Answer 12

You differentiate both the numerator and denominator separately till you get a finite value of the resulting limit on substituting 0.
So differentiate numerator and denominator separately twice.

Answer 13

\[\begin{array}{cl}
&\lim_{x\rightarrow0}\frac{1-\cos ax}{x\sin3x}\\
=&\lim_{x\rightarrow0}\frac{a\sin ax}{\sin 3x+3x\cos3x}\\
=&\lim_{x\rightarrow0}\frac{a^2\cos ax}{3\cos3x+3\cos3x-9x\sin3x}\\
=&\lim_{x\rightarrow0}\frac{a^2\cos ax}{6\cos3x-9x\sin3x}\\
=&\frac{a^2\cos0}{6\cos0-9(0)\sin0}\\
=&\frac{a^2}{6-0}\\
=&\frac{a^2}6
\end{array}\]

Answer 14

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