2
$\begingroup$

So I was solving the Cengage Mathematics Calculus book for JEE Adv. and I came across this question in the examples $$\lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}}$$ The solution given in the book uses the expansion of $\sin x$ in solving the question, whereas WolframAlpha used L'Hopital's Rule in solving it. My first instinct to solving the question when I read it was like this- $$ \lim\limits_{x \to 0} {\frac{\sin x-x}{x^3}} $$ $$ \implies \lim\limits_{x \to 0} \frac{\sin x}{x} \times \frac{1}{x^2} - \frac{x}{x^3} $$ $$ \implies \lim\limits_{x \to 0} \frac{\sin x}{x} \times \lim\limits_{x \to 0} \frac{1}{x^2} - \lim\limits_{x \to 0} \frac{x}{x^3} $$ $$ \implies 1 \times \lim\limits_{x \to 0} \frac{1}{x^2} - \lim\limits_{x \to 0} \frac{1}{x^2} $$ $$ \implies 1 \times \lim\limits_{x \to 0} \frac{1}{x^2} - \frac{1}{x^2} $$ $$ \implies 1 \times 0 $$ $$ \implies 0 $$

I do not understand, where I am going wrong with this procedure. I would really appreciate it if someone can answer. Thank you.

For reference here is the solution given in Cengage.Here is the solution given in my book

$\endgroup$
9
  • 4
    $\begingroup$ $\lim_{x\to 0}{1\over x^2}=+\infty.$ $\endgroup$ Dec 26, 2022 at 8:16
  • 1
    $\begingroup$ The limitation $\lim \frac{\sin x}{x^3} != \lim \frac{\sin x}{x}\lim \frac{1}{x^2}$ since the $\lim \frac{\sin x}{x^3}$ does not exist, $\endgroup$
    – Chia
    Dec 26, 2022 at 8:17
  • 5
    $\begingroup$ And none of your “implications” are actually implications. You should use the equals sign instead (except that in the very first step, the equality would be false). $\endgroup$ Dec 26, 2022 at 8:22
  • 3
    $\begingroup$ The main problem is that the equality $\lim_{x\to a} (f(x)+g(x))=\lim_{x\to a} f(x) + \lim_{x\to a} g(x)$ is true when both the limits $\lim_{x\to a} f(x)$ and $\lim_{x\to a}g(x)$ both exists . Otherwise it may fail. So your mistake lies in your very first step. $\endgroup$ Dec 26, 2022 at 8:36
  • 1
    $\begingroup$ @Mr.GandalfSauron I would add: both exist and are not infinity of different sign $\endgroup$ Dec 26, 2022 at 8:41

3 Answers 3

6
$\begingroup$

The limit $$\lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2} - \frac 1{x^2}\right)$$ is of the indeterminate form $\infty - \infty$ so $$\lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2} - \frac 1{x^2}\right) \neq \lim_{x\to 0}\left(\frac {\sin x}x \cdot \frac 1{x^2}\right) - \lim_{x\to 0}\left(\frac 1{x^2}\right)$$ because the RHS makes no sense.

$\endgroup$
1
  • 1
    $\begingroup$ Yes it makes sense to me now, after @Mr.GandalfSauron pointed out that the limit doesn't exist, it almost clicked for me and seeing your explanation, showing that it is of the indeterminate form $\infty - \infty$ it fully makes sense. Thank you very much $\endgroup$ Dec 26, 2022 at 10:23
2
$\begingroup$

Proper evaluation ultimately rests on the result

$\underset{x\to0}{\lim}\dfrac{\sin(x)}{x}=1,$

as methods based on differentiation (including Taylor seties) ultimately require this fact to establish the needed derivate of the sine function.

Below is a method that properly uses the aboceresilt without introducing infinite quantities or requiring differentiation:

Let

$L=\underset{x\to0}{\lim}\dfrac{\sin(x)-x}{x^3}$

and multiply in a factor of $\cos(x)$ whose limit is just $1$. Thus

$L=\underset{x\to0}{\lim}\dfrac{\sin(x)\cos(x)-x\cos(x)}{x^3}$

We apply thedoue angle sine formula and reintroduce the limit on the right side of the equation:

$L=\underset{x\to0}{\lim}\dfrac{(1/2)\sin(2x)-x\cos(x)}{x^3}$

$=\underset{x\to0}{\lim}\dfrac{(1/2)(\sin(2x)-2x)-x(-1\cos(x))}{x^3}$

$=\underset{x\to0}{\lim}\dfrac{(4)(\sin(2x)-2x)}{(2x)^3}-\underset{x\to0}{\lim}\dfrac{-1\cos(x)}{x^2}$

$=4L-\underset{x\to0}{\lim}\dfrac{1-\cos(x)}{x^2}$

Thus

$3L=-\underset{x\to0}{\lim}\dfrac{1-\cos(x)}{x^2}$

$=-(1/2)\underset{x\to0}{\lim}\dfrac{\sin^2(x/2)}{x^2}=-1/2$

where we have use the half-angle sine formula and then the $\sin(x)/x$ limit. Thereby

$L=\underset{x\to0}{\lim}\dfrac{\sin(x)-x}{x^3}=-1/6.$

$\endgroup$
3
  • 1
    $\begingroup$ Assuming the limit exists $\endgroup$ Dec 26, 2022 at 10:13
  • 2
    $\begingroup$ The question was not about an alternative to Taylor or De L'Hopital $\endgroup$ Dec 26, 2022 at 10:17
  • $\begingroup$ This is a fascinating solution, thank you for taking the time to give this solution, though my question was asking the fault in my solution, I appreciate your answer, it is very insightful. Thank you $\endgroup$ Dec 26, 2022 at 10:22
1
$\begingroup$

Assuming the limit exists, let $$l=\lim_{x \to 0}\frac{\sin x-x}{x^3} $$ $$ l=\lim_{x \to 0}\frac{\sin x-x}{x^3}\overset{x=2t}{=}\lim_{t \to 0}\frac{\sin 2t-2t}{8t^3}=\lim_{x \to 0}\frac{2\sin x \cos x-2x}{8x^3} $$ Hence $$4l=\lim_{x \to 0}\frac{\sin x \cos x-x}{x^3} $$
Now consider $$3l=4l-l=\lim_{x \to 0}\frac{\sin x \cos x-x}{x^3}-\lim_{x \to 0}\frac{\sin x-x}{x^3}=\lim_{x \to 0}\frac{\sin x \cos x-\sin x}{x^3}\\ 3l=\lim_{x \to 0}\frac{\ \sin x}{x} \lim_{x \to 0}\frac{\cos x-1}{x^2}=-\frac{1}{2} $$ Finally $$ \boxed{l=\lim_{x \to 0}\frac{\sin x-x}{x^3}=-\frac{1}{6}}$$

$\endgroup$
3
  • 2
    $\begingroup$ The question was not about an alternative to Taylor or De L'Hopital $\endgroup$ Dec 26, 2022 at 10:18
  • $\begingroup$ Thank you for the alternate solution, though it wasn't the question I asked, I appreciate your insight, it is very helpful. $\endgroup$ Dec 26, 2022 at 10:21
  • $\begingroup$ @enzotib I know, I just suggested an alternative method $\endgroup$ Dec 26, 2022 at 17:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .