Question

Find the limit

Answer 1

\(\large \color{black }{\begin{align}
&&&&&&&\lim_{x\to0}\dfrac{\left(1-\cos 2x\right)\left(3+\cos x\right)}{x\tan 4x}\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

it is same as
\(\large \color{black }{\begin{align}
&&&&&&&&\lim_{x\to0}\dfrac{\left(1-\cos 2x\right)\left(3+\cos x\right)}{4x^2}\hspace{.33em}\\~\\
\end{align}}\)

Answer 3

\[1-\cos2x=\sin ^{2}x\]

Answer 4

convert \[\tan4x=\frac{ \sin4x }{ \cos4x }\]

Answer 5

in the denominator ther is sin4x
multiply and divide by 4x in the denominator

Answer 6

i thought applying \(\lim_{x\to0} \dfrac{\tan 4x}{4x}=1\) is valid

Answer 7

s\[\lim_{x \rightarrow 0}\frac{ \sin4x }{ 4x }=1\]

Answer 8

this seems to work
\[ \lim_{x\to0}\dfrac{\left(1-\cos 2x\right)\left(3+\cos x\right)}{4x^2} \\
\lim_{x\to0}\dfrac{2\sin^2(x)}{4x^2}\left(3+\cos x\right)
\]
and use lim sinx/ x = 1

Answer 9

so this transformation is wrong


\(\large \color{black }{\begin{align}
&\lim_{x\to0}\dfrac{\left(1-\cos 2x\right)\left(3+\cos x\right)}{x\tan 4x}\hspace{.33em}\\~\\
=&\lim_{x\to0}\dfrac{\left(1-\cos 2x\right)\left(3+\cos x\right)}{4x^2}\hspace{.33em}\\~\\
\end{align}}\)

?

Answer 10

i got confused! phi is correct
yes the trnsformation is write
sorry@mathmath333

Answer 11

Yes, I can you can to that

Answer 12

so it is correct

Answer 13

thnks all

Answer 14

yes. we can use properties like lim (a/b) = lim(a)/lim(b)
to rework the expression. It is a bit tedious to type it out... but it works