How would I solve the following limits? As $$\lim_{x\to 0}x^2(1+ \cot^2{3x})$$

I know $\cot^2(3x)$ is $\displaystyle\frac{\cos^2{3x}}{\sin^2{3x}}$ but I am not sure how to solve it.

The second question is how would I solve

$$\lim_{x\to\pi/4}\frac{1-\cos x}{x}$$

  • $\begingroup$ You should try to write your mathematics with LaTeX. In the FAQ section you can find directions. $\endgroup$ – DonAntonio Jan 28 '13 at 3:17
  • $\begingroup$ Let me know, Fernando, if my answer to you last limit question was helpful, and if so, would you like to accept it? $\endgroup$ – amWhy Jan 28 '13 at 4:38

If I am reading your question right, we want $$\lim_{x\to 0}x^2(1+\cot^2 3x).$$ How to approach? A reasonable thing to do is to express the cotangent in terms of the more familiar sines and cosines: $\cot t=\dfrac{\cos t}{\sin t}$. A little bit of algebra shows that we want $$\lim_{x\to 0} \frac{x^2}{\sin^2 3x}(1+\cos^2 3x).$$

The $1+\cos^2 3x$ part is "safe," it approaches $2$.

To deal with the rest, express it as $\dfrac{1}{9}\left(\dfrac{3x}{\sin 3x}\right)^2$. Now you can take over.

In the second question, we want $$\lim_{x\to\pi/4}\frac{1-\cos x}{x}.$$ This one requires almost no work. As $x$ approaches $\pi/4$, top and bottom behave perfectly nicely. The top approaches $1-\cos(\pi/4)$, since cosine is continuous. And of course the bottom approaches $\pi/4$. For a more explicit answer, recall that $\cos(\pi/4)=1/\sqrt{2}$. So the limit is equal to $$\frac{1-\frac{1}{\sqrt{2}}}{\pi/4}.$$

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  • $\begingroup$ Sorry on the second one I meant as x approaches pi/4 $\endgroup$ – Fernando Martinez Jan 28 '13 at 3:15
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    $\begingroup$ How is $$x^2(1+\cot^2(3x))=\frac{x^2}{\sin^2(3x)}(1+\cos^2(3x))$$? $\endgroup$ – Aang Jan 28 '13 at 3:15
  • $\begingroup$ so on the second problem what would be the final answer 1-cos(pi/4) is ((2-square root(2))/2 on the denominator it is square root(2)/2 $\endgroup$ – Fernando Martinez Jan 28 '13 at 3:24

Recall that $1+\cot^2(y) = \csc^2(y)$. Hence, $$\lim_{x \to 0} x^2 (1+\cot^2(3x)) = \lim_{x \to 0} x^2 \csc^2(3x) = \lim_{x \to 0} \left(\dfrac{x}{\sin(3x)} \right)^2 = \dfrac19$$

For the second problem, as $x \to \pi/4$, the numerator tends to $1-\dfrac1{\sqrt{2}}$, while the denominator tends to $\pi/4$. Hence, the limit is $$\dfrac{1-\dfrac1{\sqrt{2}}}{\pi/4} = \dfrac{4-2\sqrt2}{\pi}$$

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  • $\begingroup$ the thing is I am asking as x approach pi/4 on the second problem $\endgroup$ – Fernando Martinez Jan 28 '13 at 3:20

$\lim_{x\to\pi/4}\frac{1-\cos x}{x} = \lim_{x\to\pi/4}\frac{2\sin^2 x/2}{x} = \lim_{x\to\pi/4}\frac{2\sin^2 x/2}{x^2 /4}(x/2) = {\pi/8}$

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  • $\begingroup$ A), I don't think this is right. B), why would you compute like this? $\endgroup$ – pjs36 Apr 15 '16 at 4:44
  • $\begingroup$ I am just use the formula,and please help improve me if i am wrong ☺ $\endgroup$ – Sokleng Apr 15 '16 at 14:38

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