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$$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}$$

Solution

\begin{align}\lim_{x\to 0} \frac{\tan x - \sin x}{x^3}&=\\&=\lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}\\ &= \lim_{x\to 0}\frac{\tan x}{x}\lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{\sin x}{x}\lim_{x\to 0} \frac{1}{x^2}\\&= \lim_{x\to 0} \frac{1}{x^2} -\lim_{x\to 0} \frac{1}{x^2}\\ &= \lim_{x\to 0} \frac{1}{x^2} -\frac{1}{x^2}\\&=0 \end{align}

But the answer is $\dfrac{1}{2}$ by L'Hopital's Rule.

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    $\begingroup$ $\infty -\infty = ?$ $\endgroup$
    – Math Lover
    Nov 21 '18 at 17:33
  • $\begingroup$ edited it, $lim_{x\to 0} \space 0 = 0$ $\endgroup$
    – Raghav
    Nov 21 '18 at 17:37
  • $\begingroup$ The same mistake applies regardless. You can’t break down a limit like that if the any of the individual limits are undefined/don’t exist. $\endgroup$
    – KM101
    Nov 21 '18 at 17:38
  • $\begingroup$ But if we have $ \lim_{x\to 1} \frac{x}{x-1} - \frac{1}{x-1}$, It will be indeterminate form if we put X=1 in it but we get 1 by adding them. $\endgroup$
    – Raghav
    Nov 21 '18 at 17:44
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    $\begingroup$ Please re-write your title to be informative. Someone else with the exact same problem would never find this solution based on its title. $\endgroup$ Nov 21 '18 at 19:36
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Your problem arises from the fact that you used $\color{red}{\lim_\limits{x \to 0} \frac{1}{x^2}}$, which does not have any finite defined value. In the end, you reach an indeterminate form $\color{red}{\infty-\infty}$...

Only split an initial limit into a product if the individual limits are defined.

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  • $\begingroup$ Can't I obtain 0 by subtracting $\frac{1}{x^2}$ by $\frac{1}{x^2}$? $\endgroup$
    – Raghav
    Nov 21 '18 at 17:39
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    $\begingroup$ You can’t break down a limit into a product if the individual limits aren’t defined. That’s where your error arose. $\endgroup$
    – KM101
    Nov 21 '18 at 17:40
  • $\begingroup$ Ohh, ok. So the the functions need to exist at the limit to be able to break down. Thank you very much man! $\endgroup$
    – Raghav
    Nov 21 '18 at 17:46
  • $\begingroup$ Exactly! (No problem.) $\endgroup$
    – KM101
    Nov 21 '18 at 17:47
  • $\begingroup$ @Raghav - The function does not need to exist, but the limit does. If two of the three limits $\lim f, \lim g, \lim (f + g)$ are known to exist (as a finite number), then it is guaranteed that the third also exists and the relationshil $$\lim f + \lim g = \lim (f+g)$$ holds. But when only one of the limits is known to exist, there is no reason that the formula has to be true. $\endgroup$ Nov 22 '18 at 2:58
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This is just another way of saying what the others told you.

$$\lim_{x\to 0} \frac{\tan x - \sin x}{x^3} \ne \lim_{x\to 0} \frac{\tan x}{x^3} - \lim_{x\to 0} \frac{\sin x}{x^3}$$

The theorem is IF $\displaystyle \lim_{x\to 0}f(x) = L$ and $\displaystyle \lim_{x\to 0}g(x)=M$, where $M, N \in \mathbb R$, THEN $\displaystyle \lim_{x\to 0}(f(x)-g(x))=L-M$

But, since $\displaystyle \lim_{x\to 0} \frac{\tan x}{x^3} = \lim_{x\to 0} \frac{\sin x}{x^3} = \infty$, then the theorem does not apply.

This limit can be evaluated without resorting to L'Hospital.

\begin{align} \frac{\tan x - \sin x}{x^3} &= \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} \\ &= \frac{\sin x - \sin x \cos x}{x^3 \cos x} \\ &= \frac{1}{\cos x} \cdot\frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2} \\ &= \frac{1}{\cos x} \cdot\frac{\sin x}{x} \cdot \frac{2\sin^2(\frac 12x)}{x^2} \\ &= \frac{1}{\cos x} \cdot\frac{\sin x}{x} \cdot \frac 12 \cdot \left(\frac{\sin \frac x2}{\frac x2}\right)^2 \\ \end{align}

which approaches $\dfrac 12$ as $x$ approaches $0$.

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    $\begingroup$ @Steven where did I do $\infty - \infty$ $\endgroup$
    – Akash Roy
    Nov 22 '18 at 13:05
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    $\begingroup$ Your way is correct. $\endgroup$
    – KM101
    Nov 22 '18 at 13:51
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    $\begingroup$ @AkashRoy Yes what you did is correct. I was looking at someone elses answer by mistake. $\endgroup$ Nov 22 '18 at 14:14
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    $\begingroup$ I didn't know we needed to go to Le Hospital to solve limits :) $\endgroup$ Nov 22 '18 at 22:36
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    $\begingroup$ @stevengregory Check your spelling of "L'Hôpital" and then re-read my comment. $\endgroup$ Nov 23 '18 at 1:07
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I don't know is there later mistakes or not, but I think there's a mistake at first equation. $ \lim\limits_{x \to 0}\big( f(x) - g(x)\big)$ is not always equal to $ \lim\limits_{x \to 0} f(x) - \lim\limits_{x \to 0} g(x)$.

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    $\begingroup$ This is the most concise correct answer. The obvious example is to pick any function $f:\mathbb R\to\mathbb R$ such that $\lim\limits_{x\to 0}$ does not exist. Then $\lim\limits_{x\to 0}\big( f(x)-f(x)\big)$ exists and equals $0$, but it is not equal to $\lim\limits_{x\to 0}f(x) + \lim\limits_{x\to 0}-f(x)$ since neither one of the latter two limits exists. $\endgroup$
    – MPW
    Nov 21 '18 at 18:30
  • $\begingroup$ There has to be further problems with the work, as splitting works when there's a finite limit. If the OP had worked through the problem correctly, they should have found that the separate limits were indeterminate. The issue is not just splitting, but then recombining them later to cancel out the indeterminate. $\endgroup$ Nov 21 '18 at 19:57
  • $\begingroup$ Is my answer okay? $\endgroup$
    – Akash Roy
    Nov 22 '18 at 8:00
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Another way of evaluation can be the use of Taylor Maclurin Expansion of $tan x$ and $sin x$.

We have

$$\lim_{x \to 0} \tan x= \frac{x}{1} +\frac{x^{3}}{3} +\frac{2x^{5}}{15} + . . .$$

$$\lim_{x \to 0} \sin x= \frac{x}{1} - \frac{x^{3}}{6} +\frac{x^{5}}{120} + . . .$$

Therefore expression turns to,

$$\lim_{x \to 0} \frac{\frac{x}{1} +\frac{x^{3}}{3} +\frac{2x^{5}}{15} + . . . - (\frac{x}{1} - \frac{x^{3}}{6} +\frac{x^{5}}{120} + . . .)}{x^{3}}$$

Cancel the $x$ and then enforce the limit after dividing the numerator by $x^{3}$ . The expression simplifies to the calculation of sum of $\frac{1}{3}$ and $\frac{1}{6}$ which is $\color{red} {\frac{1}{2}}$

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  • $\begingroup$ Cancel the x/1 term first and then divide the expression by $x^{3}$ then enforce the limit which is x tends to 0 . $\endgroup$
    – Akash Roy
    Nov 22 '18 at 5:33
  • $\begingroup$ Good solution, however I would use truncation (with big-$O$) rather than those dots. And then you don't even need the $x^5$ terms, so you can truncate at $O(x^5)$. $\endgroup$ Nov 22 '18 at 12:51
  • $\begingroup$ Ok sir Jean Claude Arbaut . Thanks for your words. $\endgroup$
    – Akash Roy
    Nov 22 '18 at 13:03
  • $\begingroup$ That's a nice solution. $\endgroup$
    – Raghav
    Nov 22 '18 at 16:24
  • $\begingroup$ Thanks @Raghav bro. $\endgroup$
    – Akash Roy
    Nov 22 '18 at 16:39
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Don't try to take the limit of each seperately & then take the difference - you just get the difference between two infinities! Express $\sin$ & $\tan$ as Taylor series - each has first term in $\theta^1$ with coefficient 1, so in the difference it drops out. If you plot $\sin\theta-\tan\theta$ it looks like a cubic at the origin. Then if you divide that series by $\theta^3$, & you get a series with an initial term in $\theta^0$, ie a constant term. (This is shown in plots: if you plot that curve just described, ÷by $\theta^3$, it begins somewhere along the y -axis instead of at the origin.) This is then all that is left as $\theta\rightarrow 0$. That's equivalent to tracing the plot I have just described in parenthesis to its point of intersection with the y -axis.

To actually get the answer immediately you just subtract the coefficient for $\theta^3$ in the series for $\sin\theta$ from that in that for $\tan\theta$, & you get 1/3 - -1/6 = 1/2.

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