$\lim\limits_{x \to 0} \frac{3x-\sin 3x}{x^3}$
I need to prove that this limit equals to $\frac{9}{2}$. Can someone give me a step by step solution?
EDIT: I am sorry. The $x$ goes to $0$, not $1$.
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ Is it supposed to be $x\to\color{red}1$? $\endgroup$– Simply Beautiful ArtFeb 28, 2017 at 23:33
-
$\begingroup$ Are you sure that you wrote the correct thing? That limit isn't 9/2. $\endgroup$– ZarosFeb 28, 2017 at 23:33
-
1$\begingroup$ math.stackexchange.com/questions/387333/… $\endgroup$– lab bhattacharjeeMar 1, 2017 at 1:54
Add a comment
|
7 Answers
$\begingroup$
$\endgroup$
3
If you allow Taylor expansions, recall that
$$\sin(x)=x-\frac16x^3+\mathcal O(x^5)$$
Thus,
$$\sin(3x)=3x-\color{red}{\frac92}x^3+\mathcal O(x^5)$$
Thus,
$$\begin{align}\frac{3x-\sin(3x)}{x^3}&=\frac{\frac92x^3+\mathcal O(x^5)}{x^3}\\&=\frac92+\mathcal O(x^2)\\&\to\frac92\end{align}$$
answered Feb 28, 2017 at 23:37
-
-
$\begingroup$ Taylor expansions make everything easy especially this limit as this limit has a question dedicated to it for proving without Taylor or L'hoptial. $\endgroup$– user312097Mar 1, 2017 at 0:02
-
$\begingroup$ @A---B Indeed, but it's not to hard to derive the Taylor expansion either. Once you know the basic trig limits, deriving derivative of derivatives is not hard at all. $\endgroup$ Mar 1, 2017 at 0:09
$\begingroup$
$\endgroup$
0
Applying L'Hopital's rule three times $$\lim_{x \to 0}\frac{3x-\sin(3x)}{x^3}=\lim_{x \to 0}\frac{3-3\cos(3x)}{3x^2}=\lim_{x \to 0}\frac{9\sin(3x)}{6x}=\lim_{x \to 0}\frac{27\cos(3x)}{6}=\frac{9}{2}.$$
$\begingroup$
$\endgroup$
1
Using l'Hôpital's rule;$$\lim_{x\to0}\frac{3x-\sin(3x)}{x^3}=\lim_{x\to0}\frac{3-3\cos(3x)}{3x^2}=\lim_{x\to0}\frac{9\sin(3x)}{6x}=\lim_{x\to0}\frac{27\cos(3x)}{6}=\frac{9}{2}$$
-
$\begingroup$ Yeah i tried to use l'Hôpital's rule but I did something wrong. Thanks for helping me out $\endgroup$– ThomasFeb 28, 2017 at 23:51
$\begingroup$
$\endgroup$
2
By elementary means:
From
$$\sin 3x=3\sin x-4\sin^3x$$
we draw
$$L=\lim_{x\to0}\frac{3x-3\sin x+4\sin^3x}{x^3}=\lim_{x\to0}\frac{3x-3\sin x}{x^3}+4.$$
But
$$\lim_{x\to0}\frac{x-\sin x}{x^3}=\lim_{3x\to0}\frac{3x-\sin3x}{27x^3}$$ so that
$$L=\frac L9+4.$$
-
1$\begingroup$ This doesn't prove that $L$ exists, does it? $\endgroup$– user1551Mar 1, 2017 at 12:43
-
$\begingroup$ @user1551: that's right, it only proves that if it exists, it has value $9/2$. $\endgroup$– user65203Mar 1, 2017 at 13:10
$\begingroup$
$\endgroup$
2
Hint: Apply the Hospital rule 3 times you obtain $\lim_{x\rightarrow 0}{{27\cos(3x)}\over 6}={9\over 2}$.
Simply Beautiful Art
74.7k12 gold badges123 silver badges281 bronze badges
answered Feb 28, 2017 at 23:37
-
7$\begingroup$ No offense, but I think this is rather low quality. $\endgroup$ Feb 28, 2017 at 23:38
-
4$\begingroup$ Why so? Not agreeing or disagreeing, but could you elaborate just a bit? $\endgroup$ Feb 28, 2017 at 23:46
$\begingroup$
$\endgroup$
3
Use L'Hospital's rule:
$$\lim _{x\to0} \frac{3x-\sin3x}{x^3} = \lim_{x\to0} \frac{3-3\cos3x}{3x^2} = \lim _{x\to0} \frac{9\sin3x}{6x} = \lim_{x\to0} \frac{27\cos3x}{6} = \frac{27}{6} = \frac{9}{2}$$
-
1$\begingroup$ There is a symbol
\toin $\LaTeX$ which renders as a right arrow: $\to$ and serves for limits instead of a 'minus'+'greater than' pair. $\endgroup$– CiaPanMar 1, 2017 at 11:24 -
$\begingroup$ I am rather new to LATEX so I am not quite acclimatized with it.Please bear with me. $\endgroup$ Mar 2, 2017 at 4:01
-
$\begingroup$ Don't worry, everyone was once a beginner :) $\endgroup$– CiaPanMar 2, 2017 at 21:13
$\begingroup$
$\endgroup$
first :
theorem :
let $f(0)=0 , f'(0)$ have existed then :
$$\lim_{x\to 0} \frac{f(x)-\sin(f(x))}{x^3}= \frac{1}{6}(f'(0))^3$$
now :
$$\lim_{x\to 0} \frac{3x-\sin(3x)}{x^3}= \frac{1}{6}(3)^3=\frac{9}{2}$$
