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lim x->0 sinx/x =
= 1
lim x->0 1-cosx/x² =
= 1/2
lim x->∞ (1+1/x)² =
= e
lim x-> -∞ (1+1/x)²=
= e
lim x->0 log(1+x)/x =
= 1
lim x-> 0 eˣ -1 / x =
= 1
lim x->∞ eˣ/xᵇ
= ∞
lim x->∞ logx/xᵇ =
= 0
lim x->0+ xᵇlogx =
= 0
d/dx logx
1/x
d/dx sinx
cosx
d/dx cosx
-sinx
d/dx tanx
sec²x
d/dx cotx
-cosec²x
d/dx arcsinx
1/√(1-x²)
d/dx arccosx
-1/√(1-x²)
d/dx arctanx
1/√(1+x²)
d/dx sinhx
coshx
d/dx coshx
sinhx
d/ dx arccoshx
1/√(x²-1)
Sequence Convergence Tests
- Null sequence test
- The comparison test
- The ratio test
- The root test
- The Integral test
- Alternating Series Test
- Absolute Convergence Test
Null Sequence Test
if Σaₙ converges, then an->0
test for divergence only: if an->0 then Σaₙ diverges
The Comparison Test
Suppose an>=0 bn>=an for n then:
- Σbₙ converges <=> Σaₙ converges
- Σaₙ diverges <=> Σbₙ diverges
The Ratio Test
Suppose Σaₙ is a series of non-negative terms and: aₙ₊₁/aₙ -> r then: 1. if r<1 => Σaₙ converges 2. if r>1 Σaₙ diverges 3. if r=0 inconclusive
The root test
Suppose Σaₙ is a series of non-negative terms and aₙ¹/ⁿ->. Then:
- if r<1 => Σaₙ converges
- if r>1 => Σaₙ diverges
- if r=0 inconclusive
Integral Test
Suppose f:[1,∞)-> is continuous, decreasing and positive in its domain. Then the series Σfₙ converges if and only if the sequence (∫f(x)dx) converges
Alternating series test
Consider the series Σ(-1)ⁿ⁺¹aₙ where the sequence (aₙ) is decreasing and converging to zero. Then the series converges.
Absolute Convergence Test
If a series converges absolutely absolutely, then it converges
