Question

I have couple of questions to ask:) please HELP!
1. lim x->0 {x-xcosx}/{sin^2(3x)}
2. lim x->1 cos^-1 (lnx)
3. lim x->0 cos {pi/sqrt(19-3sec2x)}
4. lim x->0+ [{1/x^(1/3)} - {1/(x-1)^(4/3)}]

Answer 1

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Answer 2

\[\lim_{x\to0}\frac{x-x\cos x}{\sin^23x}\]
Quite a few ways to go about this one.

Keep in mind the special trig limits:
\[\lim_{x\to0}\frac{\sin ax}{ax}=1\\
\lim_{x\to0}\frac{1-\cos ax}{ax}=0\]
So for this first limit, we're going to have to do some rewriting to make it somewhat resemble these special ones.
\[\begin{align*}\lim_{x\to0}\frac{x-\cos x}{\sin^23x}&=\lim_{x\to0}\frac{x}{\sin3x}\cdot\frac{1}{\sin 3x}\cdot\frac{1-\cos x}{1}\\
&=\lim_{x\to0}\frac{x}{\sin3x}\cdot\lim_{x\to0}\frac{1}{\sin3x}\cdot\lim_{x\to0}\frac{1-\cos x}{1}\\
&=\left(\lim_{x\to0}\frac{x}{\sin3x}\cdot\frac{3}{3}\right)\left(\lim_{x\to0}\frac{1}{\sin3x}\cdot\frac{3x}{3x}\right)\left(\lim_{x\to0}\frac{1-\cos x}{1}\cdot\frac{x}{x}\right)\\
&=\color{red}{\lim_{x\to0}\frac{x}{3(3x)}}\cdot\left(\lim_{x\to0}\frac{3x}{\sin3x}\right)^2\cdot\lim_{x\to0}\frac{1-\cos x}{x}
\end{align*}\]
The red part is the "residue" from multiplying by 1. Should be a piece of cake from here.

Answer 3

The second limit is easier. Directly substituting should do it.
\[\ln1=0~~\Rightarrow~~\cos^{-1}0=\cdots\]

Answer 4

Likewise for the third limit.
\[\sec2(0)=\sec0=1\]
so
\[\cos\left(\frac{\pi}{\sqrt{19-3}}\right)=\cos\frac{\pi}{4}=\cdots\]