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

\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}

Answer

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

Attribution
Source : Link , Question Author : lab bhattacharjee , Answer Author : RE60K

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