MHB Calculating $\displaystyle \lim_{x\to 0}$ Complex Limit

[Guest2]

$\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)}$

How do you calculate this one?

L'hopital gives me

$\displaystyle \lim_{x \to 0} \frac{2x\cos^2(3x^4)-\sin{2x}\cos^2(3x^4)}{12x^3}$

 

[Prove It]

$\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)}$

How do you calculate this one?

L'hopital gives me

$\displaystyle \lim_{x \to 0} \frac{2x\cos^2(3x^4)-\sin{2x}\cos^2(3x^4)}{12x^3}$

This is another $\displaystyle \begin{align*} \frac{0}{0} \end{align*}$ indeterminate form, so use L'Hospital's Rule again. If you keep getting indeterminate forms, keep using it...

 

[Guest2]

This is another $\displaystyle \begin{align*} \frac{0}{0} \end{align*}$ indeterminate form, so use L'Hospital's Rule again. If you keep getting indeterminate forms, keep using it...

Yeah, but this time you have to use it like four times!

I got it solved by series expansion, though, so all is good. :)

 

[Greg]

$$\begin{align*}\lim_{x\to0}\dfrac{x^2-\sin^2x}{\tan3x^4}&=\lim_{x\to0}
\dfrac{x^2-\sin^2x}{\sin3x^4}\cdot\lim_{x\to0}\cos3x^4 \\
&=\lim_{x\to0}\dfrac{2x-\sin2x}{12x^3\cos3x^4} \\
&=\lim_{x\to0}\dfrac{2x-\sin2x}{12x^3} \\
&=\lim_{x\to0}\dfrac{2-2\cos2x}{36x^2} \\
&=\lim_{x\to0}\dfrac{4\sin2x}{72x} \\
&=\lim_{x\to0}\dfrac{8\cos2x}{72} \\
&=\dfrac19\end{align*}$$

 

[Guest2]

$$\begin{align*}\lim_{x\to0}\dfrac{x^2-\sin^2x}{\tan3x^4}&=\lim_{x\to0}
\dfrac{x^2-\sin^2x}{\sin3x^4}\cdot\lim_{x\to0}\cos3x^4 \\
&=\lim_{x\to0}\dfrac{2x-\sin2x}{12x^3\cos3x^4} \\
&=\lim_{x\to0}\dfrac{2x-\sin2x}{12x^3} \\
&=\lim_{x\to0}\dfrac{2-2\cos2x}{36x^2} \\
&=\lim_{x\to0}\dfrac{4\sin2x}{72x} \\
&=\lim_{x\to0}\dfrac{8\cos2x}{72} \\
&=\dfrac19\end{align*}$$

Very nice, thank you.