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Ln (1 + 2x) divided by sin3x, X tends to 0, evaluated
x→0
Then 2x → 0
3x→0
So ln (1 + 2x) and 2x are equivalent infinitesimals
Sin3x and 3x are equivalent infinitesimals
So the original formula = LIM (x → 0) (2x) / (3x) = 2 / 3
The limit of Ln (x / x-1) - 1 / X when x tends to positive infinity
Ln [x / (x-1)] - 1 / X!
When x → + ∞, X / (x-1) → 1, then ln [x / (x-1)] → 0, and 1 / X → 0,
So, the original limit is 0
Find the limit of (aresinx / x) ^ (1 / ln (2 + x ^ 2)) (x tends to 0)
Like the title,
If the exponent is 1 / ln (2 + x ^ 2), then the exponent limit is 1 / LN2; if the base limit is 1, the result is 1
If the exponent is 1 / ln (1 + x ^ 2), then
Limit = e ^ LIM ((1 / ln (1 + x ^ 2)) · ln (aresinx / x))
=e^lim( ln(aresinx/x) / ln(1+x^2) )
=e^lim( ln(1+ aresinx/x -1) / ln(1+x^2) )
=e^lim( ( aresinx/x -1) / (x^2) )
=e^lim( ( aresinx -x) / (x³) )
Let u = aresinx, then x = sinu
When x tends to 0, UX tends to 0
Then the original limit = e ^ LIM ((u-sinu) / (Sin & # 179; U))
=e^lim( (u-sinu)/(u³) )
=e^lim( (1-cosu)/(3u²) )
=e^lim( (1/2)u²)/(3u²) )
=e^lim(1/6)
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