Is it always possible to find the limit of a function without using L'Hôpital Rule or Series Expansion?

For example,

$$\lim_{x\to0}\frac{\tan x-x}{x^3}$$

$$\lim_{x\to0}\frac{\sin x-x}{x^3}$$





  • 3
    $\begingroup$ Should this be tagged pre-calculus instead of calculus? $\endgroup$
    – robjohn
    Jun 24, 2015 at 21:51
  • 7
    $\begingroup$ The OP knows all the answers but perhaps wants to seek solutions outside analysis. $\endgroup$
    – Narasimham
    Mar 23, 2016 at 15:25
  • 1
    $\begingroup$ For $(\tan x - x)/x^3$ and $(\sin x-x)/x^3$ using only $\frac{\sin x}{x}\to1$ and not assuming in advance that the limits exist, see math.stackexchange.com/a/158134/1242. $\endgroup$ Jul 11, 2017 at 9:32

3 Answers 3


$$L_1=\lim_{x\to0}\frac{\tan x-x}{x^3}\quad L_2=\lim_{x\to0}\frac{\sin x-x}{x^3}\quad L_3=\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}\\L_4=\lim_{x\to0}\frac{e^x-x-1}{x^2}\quad L_5=\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}\quad L_6=\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$

Yes if we know beforehand the limit exists.

For $L_1$: $$L_1=\lim_{x\to0}\frac{\tan x-x}{x^3}\\ L_1=\lim_{x\to0}\frac{\tan 2x-2x}{8x^3}\\ 4L_1=\lim_{x\to0}\frac{\frac12\tan2x-x}{x^3}\\ 3L_1=\lim_{x\to0}\frac{\frac12\tan{2x}-\tan x}{x^3}\\ =\lim_{x\to0}\frac{\tan x}x\frac{\frac1{1-\tan^2x}-1}{x^2}\\ =\lim_{x\to0}\frac{(\tan x)^3}{x^3}=1\\ \large L_1=\frac13$$

For $L_2$: $$L_2=\lim_{x\to0}\frac{\sin x-x}{x^3}\\ L_2=\lim_{x\to0}\frac{\sin 2x-2x}{8x^3}\\ 4L_2=\lim_{x\to0}\frac{\frac12\sin 2x-x}{x^3}\\ 3L_2=\lim_{x\to0}\frac{\frac12\sin 2x-\sin x}{x^3} =\lim_{x\to0}\frac{\cos x-1}{x^2}\frac{\sin x}x\\ L_2=\frac13\lim_{x\to0}\frac{\cos x-1}{x^2}\\ L_2=\frac13\lim_{x\to0}\frac{\cos 2x-1}{4x^2}\\ 4L_2=\frac13\lim_{x\to0}\frac{\cos 2x-1}{x^2}\\ 3L_2=\frac13\lim_{x\to0}\frac{\cos 2x-\cos x}{x^2}\\ 3L_2=\frac13\lim_{x\to0}\frac{-2\sin^2\left(\frac x2\right)(2\cos x+1)}{x^2}\\ 3L_2=\frac13\lim_{x\to0}\frac{-2\sin^2\left(\frac x2\right)(2\cos x+1)}{x^2}\\ \large L_2=-\frac16$$

For $L_3$: $$L_3=\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}\\ L_3=\lim_{x\to0}\frac{\ln(1+2x)-2x}{4x^2}\\ 2L_3=\lim_{x\to0}\frac{\frac12\ln(1+2x)-x}{x^2}\\ L_3=\lim_{x\to0}\frac{\frac12\ln(1+2x)-\ln(1+x)}{x^2}\\ 2L_3=\lim_{x\to0}\frac{\ln(1+2x)-2\ln(1+x)}{x^2}\\ 2L_3=\lim_{x\to0}\frac{\ln\left(1-\frac{x^2}{(1+x)^2}\right)}{x^2}\\ \large L_3=-\frac12 $$

For $L_4$: $$L_4=\lim_{x\to0}\frac{e^x-x-1}{x^2}\\ 4L_4=\lim_{x\to0}\frac{e^{2x}-2x-1}{x^2}\\ 3L_4=\lim_{x\to0}\frac{e^{2x}-e^x-x}{x^2}\\ 12L_4=\lim_{x\to0}\frac{e^{4x}-e^{2x}-2x}{x^2}\\ 6L_4=\lim_{x\to0}\frac{\frac12e^{4x}-\frac12e^{2x}-x}{x^2}\\ 3L_4=\lim_{x\to0}\frac{\frac12e^{4x}-\frac32e^{2x}+e^x}{x^2}\\ 3L_4=\frac12\lim_{x\to0}\frac{e^x(e^x-1)^2(e^x+2)}{x^2}\\ \large L_4=\frac12$$

For $L_5$: $$L_5=\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}\\ 8L_5=\lim_{x\to0}\frac{\sin^{-1}2x-2x}{x^3}\\ 4L_5=\lim_{x\to0}\frac{\frac12\sin^{-1}2x-x}{x^3}\\ 3L_5=\lim_{x\to0}\frac{\frac12\sin^{-1}2x-\sin^{-1}x}{x^3}\\ 6L_5=\lim_{x\to0}\frac{\sin^{-1}2x-2\sin^{-1}x}{x^3}\\ 6L_5=\lim_{x\to0}\frac{\sin^{-1}\left(-4 x^3-2 \sqrt{1-4 x^2} \sqrt{1-x^2} x+2 x\right)}{x^3}\\ 6L_5=\lim_{x\to0}\frac{-4 x^3+2x(1- \sqrt{1-4 x^2} \sqrt{1-x^2})}{x^3}\\ 6L_5=\lim_{x\to0}-4+2\frac{(1- \sqrt{1-5 x^2+4x^4})}{x^2}\\ 6L_5=\lim_{x\to0}-4+2\frac{(1- \sqrt{1-5 x^2+4x^4})}{x^2}$$ Since you would consider binomial theorem as series expansion, if not well and good, if yes, then I'll do: Now let $\sqrt{1-5 x^2+4x^4}=\sum a_kx^k$, squaring both sides, $$1-5x^2+4x^4=a_0^2+2a_0a_1x+(2a_0a_2+a_1^2)x^2+(2a_0a_3+a_1a_2)x^3+(2a_0a_4+2a_1a_3+a_2^2)x^4+...$$ Now taking positive branch: $$a_0=1,a_1=0,a_2=-5/2,a_3=0,a_4=-9/8,...$$ So: $$6L_5=\lim_{x\to0}-4+2\frac{(1- (1-5x^2/2-9x^4/8...))}{x^2}\\\large L_5=\frac16$$

For $L_6$: $$L_6=\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}\\ 4L_6=\lim_{x\to0}\frac{\frac12\tan^{-1}2x-x}{x^3}\\ 3L_6=\lim_{x\to0}\frac{\tan^{-1}2x-2\tan^{-1}x}{2x^3}\\ 6L_6=\lim_{x\to0}\frac{\tan^{-1}\left(-\frac{2 x^3}{3 x^2+1}\right)}{x^3}\\ L_6=-\frac13$$

  • $\begingroup$ Don't you use, in your computations, the assumption that the limits exist and are finite? $\endgroup$ Dec 26, 2014 at 12:44
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    $\begingroup$ @PeterFranek yes, this method works only if you assure the limit exists $\endgroup$
    – RE60K
    Dec 26, 2014 at 16:36
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    $\begingroup$ @ADG. For $L_5$ you do not need to use any binomial formula or series, you can just multiply both numerator and denominator by $1+\sqrt{1-5x^2+4x^4}$. $\endgroup$
    – user164524
    Nov 5, 2015 at 22:50
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    $\begingroup$ This is at best a partial answer. You have answered the OP's particular problems, sure, but you haven't answered the more broad question I think OP was trying to get at: can limits of all "nice" functions be in principle solved without series expansion or L'hopital? $\endgroup$ Jul 3, 2016 at 9:13
  • 1
    $\begingroup$ Haha, you are quite sneaky... $\endgroup$ Feb 19, 2017 at 1:32

Using only trigonometric identities, in this answer, it is shown that $$ \lim_{x\to0}\frac{x-\sin(x)}{x-\tan(x)}=-\frac12\tag{1} $$ Therefore, if we subtract from $1$, we get $$ \lim_{x\to0}\frac{\tan(x)-\sin(x)}{\tan(x)-x}=\frac32\tag{2} $$ Using the limits proven geometrically in this answer, we can derive $$ \begin{align} \lim_{x\to0}\frac{\tan(x)-\sin(x)}{x^3} &=\lim_{x\to0}\frac{\tan(x)(1-\cos(x))}{x^3}\\ &=\lim_{x\to0}\frac{\tan(x)}x\frac{\sin^2(x)}{x^2}\frac1{1+\cos(x)}\\ &=\frac12\tag{3} \end{align} $$ we can divide $(3)$ by $(2)$ to get $$ \bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\tan(x)-x}{x^3}=\frac13}\tag{4} $$ and we can multiply $(1)$ by $(4)$ to get $$ \bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\sin(x)-x}{x^3}=-\frac16}\tag{5} $$ Note that $(4)$ implies $$ \begin{align} \lim_{x\to0}\frac{\tan(x)-x}{\tan^3(x)} &=\lim_{x\to0}\frac{\tan(x)-x}{x^3}\lim_{x\to0}\frac{x^3}{\tan^3(x)}\\ &=\frac13\cdot1\tag{6} \end{align} $$ Therefore, substituting $x\mapsto\tan^{-1}(x)$, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\tan^{-1}(x)-x}{x^3}=-\frac13}\tag{7} $$ Similarly, $(5)$ implies $$ \begin{align} \lim_{x\to0}\frac{\sin(x)-x}{\sin^3(x)} &=\lim_{x\to0}\frac{\sin(x)-x}{x^3}\lim_{x\to0}\frac{x^3}{\sin^3(x)}\\ &=-\frac16\cdot1\tag{8} \end{align} $$ Therefore, substituting $x\mapsto\sin^{-1}(x)$, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\sin^{-1}(x)-x}{x^3}=\frac16}\tag{9} $$

Using the Binomial Theorem, we have $$ \left(1+\frac xn\right)^n-1-x =\frac{n-1}{2n}x^2+\sum_{k=3}^n\binom{n}{k}\frac{x^k}{n^k}\tag{10} $$ and for $|x|\le1$, $$ \begin{align} \left|\sum_{k=3}^n\binom{n}{k}\frac{x^k}{n^k}\right| &=|x|^3\left|\sum_{k=3}^n\binom{n}{k}\frac{x^{k-3}}{n^k}\right|\\ &\le |x|^3\sum_{k=3}^\infty\frac1{k!}\\[6pt] &=|x|^3\left(e-\tfrac52\right)\tag{11} \end{align} $$ Combining $(10)$ and $(11)$ and taking the limit as $n\to\infty$ yields $$ \frac{e^x-1-x}{x^2}=\frac12+O(|x|)\tag{12} $$ and therefore, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{e^x-1-x}{x^2}=\frac12}\tag{13} $$ A simple corollary of $(13)$ is $$ \lim_{x\to0}\frac{e^x-1}x=1\tag{14} $$ Therefore, it follows that $$ \begin{align} \lim_{x\to0}\frac{e^x-1-x}{(e^x-1)^2} &=\lim_{x\to0}\frac{e^x-1-x}{x^2}\lim_{x\to0}\frac{x^2}{(e^x-1)^2}\\ &=\frac12\tag{15} \end{align} $$ If we substitute $x\mapsto\log(1+x)$ in $(15)$, we get $$ \lim_{x\to0}\frac{x-\log(1+x)}{x^2}=\frac12\tag{16} $$ Therefore, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\log(1+x)-x}{x^2}=-\frac12}\tag{17} $$

  • 7
    $\begingroup$ Why the downvote? I thought an approach that did not assume the limits exist and didn't use derivatives would be good (such as for presentation to a pre-calculus class). $\endgroup$
    – robjohn
    Jun 24, 2015 at 14:22
  • 16
    $\begingroup$ i wish users of MSE learn good habits like "do not downvote without giving a comment". This is especially more relevant to a well written answer like this. Anyway I like the solutions to limit problems which avoid the use of derivative. +1 from my end. Also the derivation of $(17)$ from $(13)$ was a new one for me. $\endgroup$
    – Paramanand Singh
    Jun 25, 2015 at 10:11

In general, $ \lim_{x \to 0} \frac{f(x) - \sum_{k = 1}^{n - 1} \frac{f^{(k)}(0)\cdot x^k}{k!}}{x^n} = \frac{f^{(n)}(0)}{n!} $. This can be proven using the Mean Value Theorem $n$ times and induction.

  • 10
    $\begingroup$ This is, of course, a form of l'Hôpital's theorem. $\endgroup$
    – egreg
    May 13, 2013 at 23:06
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    $\begingroup$ The Mean Value Theorem does not requires l'Hôpital's rule to prove, nor vice-versa for most cases where the limit is at a real value (as opposed to infinity). $\endgroup$
    – Jon Claus
    May 13, 2013 at 23:12
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    $\begingroup$ What I mean is that this case of l'Hôpital's theorem is easily proved using the mean value theorem. So, applying this case can really be "without l'Hôpital or Taylor expansion"? Your assertion is mostly the same as Taylor expansion, I believe. Anyway, the question is not well posed. To me, applying $\lim_{x\to0}(\sin x)/x=1$ is just the same as using the derivative of $\sin$, so l'Hôpital or Taylor. $\endgroup$
    – egreg
    May 13, 2013 at 23:19
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    $\begingroup$ No, that is circular logic. You need that limit before you can even compute $ \frac{d}{dx} \sin x $, which is needed for all three things you listed. It is most certainly not that same thing; in fact, it's not even a subject of calculus so much as precalculus and basic limits. $\endgroup$
    – Jon Claus
    May 13, 2013 at 23:48
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    $\begingroup$ @MrPink : The region in the upper left has a circular arc as one side. This is not a triangle. Replacing it with a triangle is only valid in the limit that $\Delta \theta \rightarrow 0$. Of course, then you are right back to the limit I specified in my challenge comment. That limit is hiding in the zoom from 0:48 to 0:51. $\endgroup$ Aug 25, 2020 at 23:48

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