Some Simple Equivalent Infinitesimals

When $x \to 0$:

\[x \sim \sin x \sim \arcsin x \sim \tan x \sim \arctan x\] \[x \sim (e^x - 1) \sim \ln(1+x)\] \[x \sim \ln(x+\sqrt{1+x^2})\] \[(1-\cos x) \sim \frac{1}{2}x^2\] \[\log_a(1+x) \sim \frac{x}{\ln a}\] \[(x-\sin x) \sim \frac{1}{6}x^3 \sim (\arcsin x - x)\] \[(\tan x - x) \sim \frac{1}{3}x^3 \sim (x-\arctan x)\] \[(\tan x - \sin x) \sim \frac{1}{2}x^3\] \[a^x-1 \sim x\ln a\] \[(1+ax)^b - 1 \sim abx\] \[\left(\sqrt[n]{1+x}-1\right) \sim \frac{x}{n}\]

Legal Cases:

• Multiplication: If you have a product of two infinitesimals, you can replace one of them with an equivalent infinitesimal.

• Division: If you have a quotient of two infinitesimals, you can replace either the numerator or the denominator (but not both) with an equivalent infinitesimal.

Important Considerations:

• Addition/Subtraction: Replacing infinitesimals in sums or differences is not generally valid.

• Composition: Replacing infinitesimals within more complex functions (like compositions) can be tricky and may not always be valid.

In cases of addition/subtraction in the numerator, replacing equivalent infinitesimals is generally not legal. This is because equivalent infinitesimals only guarantee the same limiting behavior when multiplied or divided, not when added or subtracted.

However, there are specific exceptions where it might be justifiable, but with careful consideration:

Linear Combinations: If the numerator consists of a linear combination of equivalent infinitesimals, and the denominator has the same power as the highest power of the infinitesimals in the numerator, then replacement might be valid. This is because the linear combination can be factored, effectively reducing the problem to multiplication or division of equivalent infinitesimals.

Taylor Series Approximations: If the infinitesimals are part of a Taylor series expansion, and the higher-order terms are negligible compared to the first-order term, then replacing equivalent infinitesimals in the numerator might be acceptable. This is because the higher-order terms contribute negligibly to the overall limit.