98. Cheatsheet

98.1. Limits

Epsilon-Delta

For every distance 𝜀 around 𝐿, there’s a 𝛿-range around 𝑎 that keeps 𝑓(𝑥) within 𝜀 of 𝐿.

lim𝑥→𝑥0𝑓(𝑥)=𝐿⟺∀𝜀>0,∃𝛿>0s.t.0<|𝑥−𝑥0|<𝛿⇒|𝑓(𝑥)−𝐿|<𝜀

The limit of 𝑓(𝑥) as 𝑥 approaches 𝑥0 equals 𝐿 if and only if, for every 𝜀>0, there exists a 𝛿>0 such that, whenever 0<|𝑥−𝑥0|<𝛿, implies that |𝑓(𝑥)−𝐿|<𝜀

Limit TypeNameQuantifiers
𝑥→𝑥0,𝑓(𝑥)→𝐿Epsilon-Delta∀𝜀,∃𝛿
𝑥→𝑥0,𝑓(𝑥)→∞M-Delta∀𝑀,∃𝛿
𝑥→∞,𝑓(𝑥)→𝐿epsilon-N∀𝜀,∃𝑁
𝑥→∞,𝑓(𝑥)→∞M-N∀𝑀,∃𝑁

Finite → Finite (Epsilon-Delta)

lim𝑥→𝑥0𝑓(𝑥)=𝐿⟺∀𝜀>0,∃𝛿>0s.t.0<|𝑥−𝑥0|<𝛿⇒|𝑓(𝑥)−𝐿|<𝜀

Finite → Infinity (M-Delta)

+∞

lim𝑥→𝑥0𝑓(𝑥)=+∞⟺∀𝑀>0,∃𝛿>0s.t.0<|𝑥−𝑥0|<𝛿⇒𝑓(𝑥)>𝑀

−∞

lim𝑥→𝑥0𝑓(𝑥)=−∞⟺∀𝑀>0,∃𝛿>0s.t.0<|𝑥−𝑥0|<𝛿⇒𝑓(𝑥)<−𝑀

Infinity → Finite (Epsilon-N)

+∞

lim𝑥→+∞𝑓(𝑥)=𝐿⟺∀𝜀>0,∃𝑁>0s.t.𝑥>𝑁⇒|𝑓(𝑥)−𝐿|<𝜀

−∞

lim𝑥→+∞𝑓(𝑥)=𝐿⟺∀𝜀>0,∃𝑁>0s.t.𝑥<−𝑁⇒|𝑓(𝑥)−𝐿|<𝜀

Infinity → Infinity (M-N)

lim𝑥→±∞𝑓(𝑥)=±∞⟺∀𝑀>0,∃𝑁>0s.t.𝑥<𝑁⇒𝑓(𝑥)>𝑀

98.2. Derivatives

𝑑𝑓𝑑𝑥(𝑥)=𝑓′(𝑥)=limℎ→0𝑓(𝑥+ℎ)−𝑓(𝑥)ℎ
Example

Let 𝑓(𝑥)=𝑥2

a. Find 𝑓′(𝑥)

𝑓′(𝑥)=2x

b. Prove a

𝑓′(𝑥)=limℎ→0𝑓(𝑥+ℎ)−𝑓(𝑥)(𝑥+ℎ)−𝑥=limℎ→0𝑓(𝑥+ℎ)−𝑓(𝑥)ℎ=limℎ→0(𝑥+ℎ)2−𝑥2ℎ=limℎ→0𝑥2+2𝑥ℎ+ℎ2−𝑥2ℎ=limℎ→0ℎ(2𝑥+ℎ)ℎ=limℎ→0(2𝑥+ℎ)=limℎ→0(2𝑥+0)=2x

c. Prove b

lim𝑥→𝑥0𝑓(𝑥)=𝐿⟺∀𝜀>0,∃𝛿>0s.t.0<|𝑥−𝑥0|<𝛿⇒|𝑓(𝑥)−𝐿|<𝜀

p.f.:

Let 𝜀>0

Choose 𝛿=𝜀

Suppose 0<|ℎ−0|<𝛿

Check

|(𝑥+ℎ)2−𝑥2ℎ−2𝑥|=|𝑥2+2𝑥ℎ+ℎ2−𝑥2ℎ−2𝑥|=|ℎ(2𝑥+ℎ)ℎ−2𝑥|=|2𝑥+ℎ−2𝑥|=|ℎ|<𝛿=𝜀

98.3. Integrals

98.4. Differential

If you have a function:

𝑦=𝑓(𝑥)

Then the differential d𝑢 is defined as:

d𝑦=𝑓′(𝑥)⋅d𝑥

This means:

So:

d𝑦=d𝑦d𝑥⋅d𝑥
Example

Let’s say“

𝑦=𝑥2

Then:

d𝑦d𝑥=2𝑥
𝑑d𝑥(∫𝑓(𝑥)d𝑥)=𝑓(𝑥)∫(𝑑d𝑥𝑓(𝑥))d𝑥=𝑓(𝑥)


OperationNotationInputOutputMeaning
Derivative𝑑𝑓𝑑𝑥Function𝑓(𝑥)Function𝑓′(𝑥)Slope/Rate of change
at each 𝑥
Indefinite Integral∫𝑓(𝑥)𝑑𝑥Function𝑓(𝑥)Family of functions𝐹(𝑥)+𝐶Function whose slope
is 𝑓(𝑥)
Definite Integral∫𝑎𝑏𝑓(𝑥)𝑑𝑥Function𝑓(𝑥)Bounds[𝑎,𝑏]NumberTotal signed area
between 𝑎 and 𝑏


Rule𝑑d𝑥 Rule𝑑d𝑥 Example∫ Rule∫ Example
Constant𝑑d𝑥[𝑐]=0𝑑d𝑥[7]=0∫𝑐d𝑥=𝑐𝑥+𝐶∫7d𝑥=7𝑥+𝐶
Power𝑑d𝑥[𝑥𝑛]=𝑛𝑥𝑛−1𝑑d𝑥[𝑥4]=4𝑥3∫𝑥𝑛d𝑥=𝑥𝑛+1𝑛+1+𝐶(𝑛≠−1)∫𝑥3d𝑥=𝑥44+𝐶
Constant Multiple𝑑d𝑥[𝑐𝑓]=𝑐𝑓′𝑑d𝑥[3𝑥2]=3⋅2𝑥=6𝑥∫𝑐𝑓d𝑥=𝑐∫𝑓d𝑥∫3𝑥2d𝑥=3⋅𝑥33=𝑥3+𝐶
Sum𝑑𝑑𝑥[𝑓+𝑔]=𝑓′+𝑔′𝑑d𝑥(𝑥2+𝑥)=𝑑d𝑥(𝑥2)+𝑑d𝑥(𝑥)=2𝑥+1∫(𝑓+𝑔)d𝑥=∫𝑓d𝑥+∫𝑔d𝑥∫(𝑥2+𝑥)d𝑥=∫𝑥2d𝑥+∫𝑥d𝑥=𝑥33+322+𝐶
Difference𝑑𝑑𝑥[𝑓−𝑔]=𝑓′−𝑔′𝑑d𝑥(𝑥2−𝑥)=𝑑d𝑥(𝑥2)−𝑑d𝑥(𝑥)=2𝑥−1∫(𝑓−𝑔)d𝑥=∫𝑓d𝑥−∫𝑔d𝑥∫(𝑥2−𝑥)d𝑥=∫𝑥2d𝑥−∫𝑥d𝑥=𝑥33−𝑥22+𝐶
Product𝑑𝑑𝑥[𝑓𝑔]=𝑓′𝑔+𝑓𝑔′𝑑d𝑥[𝑥sin𝑥]=1⋅sin𝑥+𝑥⋅cos𝑥

Integration by Parts

∫𝑢d𝑣=𝑢𝑣−∫𝑣d𝑢
Quotient𝑑𝑑𝑥[𝑓𝑔]=𝑓′𝑔−𝑓𝑔′𝑔2𝑑d𝑥[𝑥2𝑥+1]=2𝑥(𝑥+1)−𝑥2(1)(𝑥+1)2=𝑥2+2𝑥(𝑥+1)2Algebraic Manipulation / Substitution
Chain𝑑𝑑𝑥[𝑓(𝑔(𝑥))]=𝑓′(𝑔(𝑥))⋅𝑔′(𝑥)𝑑d𝑥[sin(𝑥2)]=cos(𝑥2)⋅2𝑥

Integration by Substitution

∫𝑓(𝑔(𝑥))𝑔′(𝑥)d𝑥=∫𝑓(𝑢)d𝑢
Exponential𝑑d𝑥[𝑒𝑥]=𝑒𝑥𝑑d𝑥[𝑒𝑥]=𝑒𝑥∫𝑒𝑥d𝑥=𝑒𝑥+𝐶∫𝑒𝑥d𝑥=𝑒𝑥+𝐶
𝑑𝑑𝑥[𝑎𝑥]=𝑎𝑥ln(𝑎)𝑑d𝑥[2𝑥]=2𝑥ln2∫𝑎𝑥𝑑𝑥=𝑎𝑥ln𝑎+𝐶∫2𝑥d𝑥=2𝑥ln2+𝐶
Logarithmic𝑑𝑑𝑥[ln(𝑥)]=1𝑥𝑑𝑑𝑥[ln(𝑥)]=1𝑥∫ln(𝑥)d𝑥=𝑥ln𝑥−𝑥+𝐶∫ln(𝑥)d𝑥=𝑥ln𝑥−𝑥+𝐶
𝑑𝑑𝑥[log𝑎(𝑥)]=1𝑥ln(𝑎)𝑑d𝑥log2(𝑥)=1𝑥ln(2)∫log𝑎(𝑥)d𝑥==𝑥ln𝑥ln𝑎−𝑥ln𝑎+𝐶∫log10(𝑥)d𝑥=𝑥ln𝑥ln10−𝑥ln10+𝐶
Sin𝑑𝑑𝑥[sin(𝑥)]=cos(𝑥)𝑑𝑑𝑥[sin(𝑥)]=cos(𝑥)∫sin(𝑥)d𝑥=−cos𝑥+𝐶∫sin(𝑥)d𝑥=−cos𝑥+𝐶
Cos𝑑𝑑𝑥[cos(𝑥)]=−sin(𝑥)𝑑𝑑𝑥[cos(𝑥)]=−sin(𝑥)∫cos(𝑥)d𝑥=sin𝑥+𝐶∫cos(𝑥)d𝑥=sin𝑥+𝐶
Tan𝑑𝑑𝑥[tan(𝑥)]=sec2(𝑥)𝑑𝑑𝑥[tan(𝑥)]=sec2(𝑥)∫tan(𝑥)d𝑥=−ln|cos𝑥|+𝐶∫tan(𝑥)d𝑥=−ln|cos𝑥|+𝐶

98.5. Product Rule ⟶ Integration by Parts

∫𝑢d𝑣=𝑢𝑣−∫𝑣d𝑢
Example

Given a function:

𝑓(𝑥)=𝑥⋅𝑒𝑥

Integrate (by parts):

∫𝑥⋅𝑒𝑥d𝑥


Step 1: Choose 𝑢 and d𝑣

We choose:

  • 𝑢=𝑥 (easy to differentiate)

  • d𝑣=𝑒𝑥 (easy to integrate)


Step 2: Compute d𝑢 and 𝑣

  • 𝑢=𝑥⟶d𝑢=d𝑥

  • d𝑣=𝑒𝑥d𝑥⟶𝑣=∫𝑒𝑥d𝑥=𝑒𝑥


Step 3: Plug into formula

∫𝑥⋅𝑒𝑥d𝑥=𝑢⋅𝑣−∫𝑣⋅d𝑢=𝑥⋅𝑒𝑥d𝑥−∫𝑒𝑥⋅d𝑥


Step 4: Compute the remaining integral:

∫𝑒𝑥d𝑥=𝑒𝑥


Step 5: Finish the expression:

∫𝑥⋅𝑒𝑥=𝑥⋅𝑒𝑥−𝑒𝑥+𝐶=𝑒𝑥(𝑥−1)+𝐶

98.6. Chain Rule ⟶ 𝑢-Substitution

∫𝑓(𝑔(𝑥))𝑔′(𝑥)d𝑥=∫𝑓(𝑢)d𝑢

98.7. Quotient Rule ⟶ Algebraic Manipulation / Substitution