## How to find the limit of LIM (x → 0) tan3x / x?

lim(x→0)tan3x/x=lim(x→0)sin3x/3x(3/cos3x)=lim(x→0)sin3x/3x[lim(x→0)(3/cos3x)]

=1x3=3

### Prove that the limit lim|x| / X does not exist (x → o)

lim(x→o+)|x|/x=lim(x→o+)x/x=1

lim(x→o-)|x|/x=lim(x→o+)(-x)/x=-1

LIM (x → O +)|x| / X is not equal to LIM (x → o -) | / x, so the limit lim| / X does not exist (x → o)

### LIM (1 + x) / (1-e ^ 1 / x) proves that the limit of X → 0 does not exist Verification

It's OK without substitution

### What does "higher order" mean in higher-order infinitesimal? thank you!

The so-called higher-order infinitesimal means that if LIM (A / b) = 0, B is an infinitesimal higher-order than a, which is recorded as B = 0 (a)

### When x →∞, arctanx ~ x is proved That is to prove that arctanx is equivalent to X

When x → + ∞, arctanx - > pi / 2, when x → - ∞, arctanx - > pi / 2, LIM (x - > 0) arctanx = LIM (x - > 0) x = 0.lim (x - > 0) arctanx / x = LIM (x - > 0) [1 / (1 + x ^ 2)] / 1 = 1. Therefore, when x → 0, arctanx and X are equivalent infinitesimals. When x → 0, arctanx ~ x

### It is proved that arcsin X and X are equivalent infinitesimals It is proved that LIM (x → 0) arcsin X / x = 1, that is, it is proved that arcsin X and X are equivalent infinitesimals, Can you use lobida's law? This problem seems to be the type of 0 / 0 finding the limit

The method of proof depends on your level of knowledge and the usefulness of those conclusions

First of all, this is equivalent to the equivalence between X and SiNx. You can directly say that this is obvious. You can say that SiNx = x-x ^ 3 / 6 + O (x ^ 3), or prove the expansion of SiNx by Taylor formula, or even start by proving Taylor formula