Calculate $\lim_{x \to 0} (\cot^3(x)·x·(\ln(1+3x)-3x))$

Question

I do not seem to be be able to calculate this limit. I tired many things using l'Hospital but it gets messier. A useful limit is $\frac{\ln(1+x)}{x}=1$ as $x\to0$. But still it still does not give the desired result. Any hints? Suggestions? $$\lim_{x \to 0} (\cot^3(x)·x·(\ln(1+3x)-3x))$$

The limit seems to be $;-\cfrac92;$ , but: can you use L'Hospital's Rule? — DonAntonio, Dec 19 '17 at 22:59

score 2 · Accepted Answer · answered Dec 19 '17 at 23:01

2

$$L = \lim_{x \to 0} {x\cot^3(x)(\ln(1+3x)-3x)} = \lim_{x \to 0} {\dfrac{\ln(1+3x)-3x}{x^2}} = 9\lim_{x \to 0} {\dfrac{\ln(1+3x)-3x}{(3x)^2}}$$

Let $y = 3x$

$$\lim_{y \to 0} {\dfrac{\ln(1+y)-y}{y^2}} = \dfrac{-1}2$$

As shown here. So $L = \dfrac{-9}{2}$

answered Dec 19 '17 at 23:01

user8277998

2,666

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where does $x^2$ come from? – Erik Hambardzumyan Dec 19 '17 at 23:02
@ErikHambardzumyan Multiply and divide by $x^2$. – user8277998 Dec 19 '17 at 23:03
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i dont see what happens with $cot^3x$ – Erik Hambardzumyan Dec 19 '17 at 23:05
@ErikHambardzumyan $\lim_{x \to 0}\cos^3 x \left(\dfrac{x}{\sin x}\right)^3$. – user8277998 Dec 19 '17 at 23:06

score 2 · Answer 2 · answered Dec 19 '17 at 23:06

2

$$\cot^3(x)\cdot x\cdot (\ln(1+3x)-3x)=(x^3\cot^3(x))\cdot \frac{\ln(1+3x)-3x}{x^2}=\\=(x^3\cot^3(x))\cdot \frac{3x-\frac{9x^2}{2}+o(x^2)-3x}{x^2}=(x^3\cot^3(x))\cdot \left(-\frac{9}{2}+o(1)\right)\to1\cdot\left(-\frac{9}{2}\right)=-\frac{9}{2}$$

answered Dec 19 '17 at 23:06

user

154,566

score 1 · Answer 3 · answered Dec 19 '17 at 22:59

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$\frac {(\cos^3 x)(x)(\ln (1+3x) - 3x)}{\sin^3 x}$

You can do a taylor expansion...

$\frac { (1)(x)(+3x - \frac {(3x)^2}{2} - 3x)}{x^3}\\ -\frac {9}{2}$

answered Dec 19 '17 at 22:59

Doug M

57,877

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how about not using taylor expansion – Erik Hambardzumyan Dec 19 '17 at 23:03

DonAntonio · Answer 4 · 2017-12-19T23:14:30.967

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Using L'Hospital's rule:

$$\lim_{x\to0}\frac{\log(1+3x)-3x}{\sin^2x}\stackrel{\text{L'H}}=\lim_{x\to0}\frac{\frac3{1+3x}-3}{\sin2x}\stackrel{\text{L'H}}=\lim_{x\to0}\frac{-\frac9{(1+3x)^2}}{2\cos2x}=-\frac92$$

and then

$$\lim_{x\to0}\;\cot^3x\cdot x\cdot\left(\log(1+3x)-3x\right)=\lim_{x\to0}\,\cos^3x\cdot\frac x{\sin x}\cdot\frac{\log(1+3x)-3x}{\sin^2x}=1\cdot1\cdot\left(-\frac92\right)=-\frac92$$

edited Dec 19 '17 at 23:14

answered Dec 19 '17 at 23:06

DonAntonio

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score 0 · Answer 5 · answered Dec 19 '17 at 23:52

L'Hopital will work if you do a little housework beforehand. The expression equals

$$ \cos^3 x\cdot\frac{x^3}{\sin^3 x}\cdot\frac{x(\ln (1+3x) - 3x)}{x^3}.$$

The first two fractions $\to 1,$ so we need only study the last fraction. This equals

$$\frac{\ln (1+3x) - 3x}{x^2}.$$

Now it's fairly easy to use L'Hopital.

score -1 · Answer 6 · answered Dec 20 '17 at 00:18

-1

Without l'Hopital, Taylor expansions or calculus:

$$\begin{aligned} L&=\frac{x\left(\ln(1+3x)-3x\right)}{\tan^3(x)} \\ u&=3x \\ L&=9\left(\frac{x}{\tan x}\right)^3\frac{\ln(1+u)-u}{u^2} \\ \lim L&=9\cdot1^3\cdot-\frac12\quad \end{aligned} $$

Using the standard limits of $\lim\frac{x}{\tan x}$ and $\lim\frac{\ln(1+u)-u}{u^2}$.

answered Dec 20 '17 at 00:18

Jam

10,325

Would the downvoter care to explain their downvote? – Jam Dec 26 '17 at 21:45

Calculate $\lim_{x \to 0} (\cot^3(x)·x·(\ln(1+3x)-3x))$

6 Answers6