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

Is it always possible to find the limit of a function without using L'Hospital 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 at 10:32

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

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

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