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

For example,

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

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





Aditya Dwivedi
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lab bhattacharjee
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3 Answers3


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

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

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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, $$ \bbox[5px,border:2px solid #C0A000]{\lim_{x\to0}\frac{\log(1+x)-x}{x^2}=-\frac12}\tag{17} $$

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    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). – robjohn Jun 24 '15 at 14:22
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    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. – Paramanand Singh Jun 25 '15 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.

Jon Claus
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    This is, of course, a form of l'Hôpital's theorem. – egreg May 13 '13 at 23:06
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    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). – Jon Claus May 13 '13 at 23:12
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    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. – egreg May 13 '13 at 23:19
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    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. – Jon Claus May 13 '13 at 23:48
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    The method above uses repeated derivatives and essentially uses the proof technique of Taylor's series. I believe the OP wanted to have a technique which avoids differentiation altogether. But when we disable differentiation, it is almost impossible to evaluate the above limits. In some of the cases if we assume the existence of the limit then we can evaluate the limit by simple algebraic/trigonometric manipulation. But for existence we need differentiation. – Paramanand Singh Jul 07 '13 at 05:43
  • The equivalence I cited does not relay on Taylor Series and can be proved with only the Mean Value Theorem for derivatives. But yes, it does rely on differentation. The OP only said he doesn't want Taylor Series or l'Hospital's rule applied. – Jon Claus Jul 08 '13 at 02:32
  • I have to disagree. Actually you don't need to know that $\lim_{x\to 0}\frac{\sin x}{x}=1$ before computing $\frac{d}{dx}\sin x$. – Mr Pink Jun 10 '20 at 23:52
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    @MrPink : Feel free to demonstrate how to evaluate $\lim_{h \rightarrow 0} \frac{\sin(x+h) - \sin(x)}{h}$ without the use of that limit (and without the use of an equivalent limit). – Eric Towers Aug 25 '20 at 17:45
  • @EricTowers Here is a proof that $\frac{d}{dx}\sin x=\cos x$ which doesn't use $\lim_{x\to 0}\frac{\sin x}{x}=1$: https://www.youtube.com/watch?v=R4o7sraVMZg – Mr Pink Aug 25 '20 at 20:00
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    @MrPink : False. That limit is used to pretend that the handwavy "not actually a triangle" is similar to the main right triangle (around 1:39). These are not similar triangles since one of these is not a triangle. Replacing the curved edge of the not-a-triangle with the tangent line to the circle can be justified in the limit, but is not justified in the video. – Eric Towers Aug 25 '20 at 20:08
  • @EricTowers Even though they use the similarity property in the video, it's actually not needed, because you can notice that the top left angle on the triangle is $\theta$. – Mr Pink Aug 25 '20 at 21:42
  • @MrPink : The similarity of the triangle with the triangle that the not-triangle approximates in the limit is crucial to the argument. *Both* $\theta$s must be interior angles of actual triangles. The not-a-triangle is not a triangle, it requires use of a limit equivalent to the one specified in my comment to replace it with an actual triangle. And only then does the similarity argument work. – Eric Towers Aug 25 '20 at 22:30
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    @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. – Eric Towers Aug 25 '20 at 23:48
  • @EricTowers Can you explain where exactly $\lim_{x\to 0}\frac{\sin x}{x}$ "pops out"? Yes, I understand that it is not a triangle, but you can see that it "approaches" a triangle. Still, I don't see where that particular sine limit "is used" in the argument at all. – Mr Pink Aug 28 '20 at 00:39
  • @MrPink : You say it "approaches" a triangle. It does so in the limit $\Delta \theta \rightarrow 0$. That is, this approximate geometric construction is computing the average rate of change of sine between $\theta$ and $\theta + \Delta \theta$ in the limit as $\Delta \theta \rightarrow 0$. Call "$\theta$" by the name "$x$" and call "$\Delta \theta$" by the name "$h$" and you have exactly the limit I wrote. – Eric Towers Aug 28 '20 at 19:30
  • @EricTowers I understand that it approaches a triangle in the limit $\Delta \theta\to 0$, but why do you have to take the sine of that angle? – Mr Pink Aug 29 '20 at 00:28
  • @MrPink : The choice of which leg of the resulting limit triangle is used in the ratio, chooses the sine. If we chose the other leg, we would be discussing cosine. – Eric Towers Aug 30 '20 at 14:38
  • @EricTowers Doesn't the fact that $f(x)=\sqrt{1-x^2}$ is differentiable there prove that the circular curve can be replaced by the line segment as $\Delta \theta\to 0$? – Mr Pink Sep 03 '20 at 15:19
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    @MrPink : "Replaced". So now you're making my argument -- Actually evaluating the rate of change in sine for a finite angle is too hard, so we take the limit, pretending that the function is adequately approximated by the tangent. – Eric Towers Sep 04 '20 at 12:12
  • @EricTowers Proving the differentiability of $f(x)=\sqrt{1-x^2}$ justifies that the arc can be replaced by a line segment. Where did that use $\lim_{\theta \to 0}\frac{\sin\theta}{\theta}=1$? – Mr Pink Sep 05 '20 at 10:13
  • @MrPink : "by **a** line segment". It is not enough to replace by **a** line segment. It is required that the line segment produce a right triangle having the correct relation between a specific acute angle and its opposite. – Eric Towers Sep 06 '20 at 02:59
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    @MrPink : Also, it appears that you are unwilling or unable to recognize that your original claim is wrong and the video directly uses the limit under question. Consequently, you are wasting our time. I won't have any more of my time wasted by you. Perhaps you will find someone willing to re-explain and re-explain this to you until you finally arrive at the correct conclusion. I won't be spending that time. – Eric Towers Sep 06 '20 at 03:00
  • @MrPink : The last I will write about this (from August 25 at 17:45) : "... demonstrate how to evaluate $\lim_{h \rightarrow 0} \frac{\sin(x+h)−\sin(x)}{h}$ without the use of that limit (and without the use of an equivalent limit).". \begin{align*} \lim_{h \rightarrow 0} \frac{\sin(x+h)−\sin(x)}{h} &= \lim_{h \rightarrow 0} \frac{\sin(x)\cos(h)+\cos(x)\sin(h)−\sin(x)}{h} \\ &= \sin(x) \lim_{h \rightarrow 0}\frac{\cos(h) -1}{h} + \cos(x) \lim_{h \rightarrow 0} \frac{\sin(h)}{h} \text{,}\end{align*} so the use of either limit to justify linearization is the use of an equivalent limit. – Eric Towers Sep 06 '20 at 05:11
  • @EricTowers Any line segment there would produce a right triangle, because any tangent line to the circle would be perpendicular to the ray emanating from the origin and going through the point of tangency. – Mr Pink Sep 06 '20 at 09:10
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    @EricTowers And the more $\Delta\theta$ gets close to $0$, the more the 'not-a-triangle' gets close to a triangle. – Mr Pink Sep 06 '20 at 09:50
  • @EricTowers I think I finally understand it: It can be proved that $\frac{d}{dx}\sin x=\cos x$ using the fact that tangent line is perpendicular to radius and that $f(x)=\sqrt{1-x^2}$ "flattens out" when zooming closer and closer (because it is differentiable). *Then*, the fact that $\frac{d}{dx}\sin x=\cos x$ implies that $\lim_{x\to 0}\frac{\sin x}{x}=1$, because, as you say, $\lim_{h\to 0}\frac{\sin (x+h)-\sin x}{h}=\sin x\,\lim_{h\to 0}\frac{\cos h-1}{h}+\cos x\, \lim_{h\to 0}\frac{\sin h}{h}$. Still, this doesn't mean that you have to compute the sin x/x limit before the derivative of sin – Mr Pink Sep 06 '20 at 10:30
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    @MrPink : The tangent line is irrelevant. The point on the circle is never on the tangent line. Continuing to think you have proven something using the tangent line is continuing to be wrong. The limit I wrote contains the unavoidable ingredients to get the **correct** point actually on the circle. Until your process uses points **on** the circle, your process is wrong. – Eric Towers Sep 07 '20 at 02:58
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    @EricTowers It's never on the tangent line, but it can get arbitrarily near. That's what the derivatives are about. Why shouldn't this reasoning work? – Mr Pink Sep 07 '20 at 04:30
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    @MrPink : Point not on circle = wrong. – Eric Towers Sep 07 '20 at 04:31
  • @EricTowers But **why** is that wrong when the error can be made arbitrarily small? – Mr Pink Sep 07 '20 at 05:24
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    @MrPink : You can't find the error until you can compute it correctly. You can't compute it correctly if the point is not on the circle. You are of course free to solve the wrong problem, using points not on the circle. While doing so may coincidentally give the correct answer, trusting a garbage result is wrong. – Eric Towers Sep 07 '20 at 05:26
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    @Mr Pink: Eric Towers is actually correct here. You've severely misunderstood the point of the video. The video isn't a rigorous proof, but rather, a visualization to help build a geometric intuition for why the limit formula for the derivative works for the sine function. In order for the asymptotic triangular similarity between the two regions in the video to be established, one needs to first perform a linearization on the point of tangency of the circle, and then actually prove that the angle between the tangent line and the horizontal axis is the angle that was claimed to be. – Angel Nov 22 '20 at 07:55
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    @Mr Pink: Also, while lim sin(θ)/θ (θ —> 0) was never explicitly referred to in the video, it is definitely implicit in the calculation that was shown in the video when appealing to the similar triangles. Notice that by definition, Δy = sin(θ + Δθ) – sin(θ), so they ultimately just used the limit definition of the derivative in abbreviated form. Additionally, the reason the visualization works is because they are able to assert that the triangles are similar in the limit, but this assertion is actually equivalent to lim sin(θ)/θ (θ —> 0) = 1 to begin with. – Angel Nov 22 '20 at 08:03
  • @Mr Pink: In other words, they appealed to the same argument that is normally used to prove lim sin(Δθ)/Δθ (Δθ —> 0), except they just an arbitrary angle θ for the tangency point instead of the special case θ = 0. In essence, the video just visually demonstrates the above limit, and then presents an arbitrarily phase shifted and vertically shifted version of it. They still had use the argument to visually demonstrate said limit to demonstrate the more general version for the derivative. – Angel Nov 22 '20 at 08:09
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    @Angel "one needs to first perform a linearization on the point of tangency of the circle" Yes. And that is not a problem at all as one can prove that $f(x)=\sqrt{1-x^2}$ is differentiable, hence it represents a line locally as $\Delta \theta\to 0$. That way, you can prove that $\frac{d}{d\theta}\sin\theta =\cos\theta$ using the argument with triangles and then, using the limit definition of the derivative, you can prove $\lim_{\theta\to 0}\frac{\sin\theta}{\theta}=1$. – Mr Pink Apr 20 '21 at 19:01
  • @EricTowers You can approximate the circle by regular $n$-gon line segments if you will, and crank up $n$ to infinity. Just observe the construction as $n$ gets bigger and bigger. – Mr Pink Apr 20 '21 at 19:12
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    @MrPink : This also does not correct the errors you have made: Essentially every choice of $n$ segments fails to produce a line perpendicular to the radius, so is *not* a tangent line. Additionally, the only points in your polygon *on* the circle are the endpoints of the segments, so the "error" you are calculating is for a point not on the circle. It has twice been patiently explained to you that your original argument is wrong. Moving the goalposts to another wrong argument is a waste of everyone's time again; I won't be participating further. – Eric Towers Apr 20 '21 at 19:21
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    @EricTowers The $n$-gon thing doesn't work here, OK. But still, one can use the differentiability of $f(x)=\sqrt{1-x^2}$. The argument is not wrong, you can see nicomezi's comment here: https://math.stackexchange.com/questions/3307345/geometric-proof-for-the-derivative-of-sine – Mr Pink Apr 20 '21 at 20:06
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    @MrPink You certainly did ignore the rest of my comment and decided to cherry-pick a quote of out its context, huh. – Angel Oct 19 '21 at 12:04