anonymous
  • anonymous
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)}]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
goformit100
  • goformit100
"Welcome to OpenStudy. I can guide regarding this useful site; ask your doubts from me, for it you can message me. Please use the chat for off topic questions. And remember to give the helper a medal, by clicking on "Best Answer". We follow a code of conduct, ( http://openstudy.com/code-of-conduct ). Please take a moment to read it."
anonymous
  • anonymous
\[\lim_{x\to0}\frac{x-x\cos x}{\sin^23x}\] 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.
anonymous
  • anonymous
The second limit is easier. Directly substituting should do it. \[\ln1=0~~\Rightarrow~~\cos^{-1}0=\cdots\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.