# Are all limits solvable without L'Hôpital Rule or Series Expansion

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

For example,

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

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

$$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$

$$\lim_{x\to0}\frac{e^x-x-1}{x^2}$$

$$\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}$$

$$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$

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edited with L5 and L6 – ADG Jan 11 '15 at 10:32
Should this be tagged pre-calculus instead of calculus? – robjohn Jun 24 '15 at 21:51
The OP knows all the answers but perhaps wants to seek solutions outside analysis. – Narasimham Mar 23 at 15:25

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

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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@lab bhattacharjee see the edit. – ADG Sep 19 '14 at 3:36
Don't you use, in your computations, the assumption that the limits exist and are finite? – Peter Franek Dec 26 '14 at 12:44
@PeterFranek yes, this method works only if you assure the limit exists – ADG Dec 26 '14 at 16:36
@ADG. For $L_5$ you do not need to use any binomial formula or series, you can just multiply both numerator and denominator by $1+\sqrt{1-5x^2+4x^4}$. – Mathematician171 Nov 5 '15 at 22:50

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.

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This is, of course, a form of l'Hôpital's theorem. – egreg May 13 '13 at 23:06
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
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
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
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 7 '13 at 5:43

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
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