Blog
Lemma
- August 16, 2020
- Category: Lemma
☼ Lemma :
Limit of a Function
- Neighbourhood of a = ( b, c ) ; Where a ∈ ( b, c )
- δ – Neighbourhood of a= N(a ,δ)
= (a-δ,a+δ)
= { x / a – δ < x < a + δ ; x ∈ R }
= { x / |x-a| < δ ; x ∈ R }
= { x / 0 ≤ |x-a| < δ ; x ∈ R }
- Deleted δ – Neighbourhood of a
Punctured δ – Neighbourhood of a = N*(a ,δ)
= N(a ,δ) – {a}
= (a-δ,a+δ) – {a}
= { x / a – δ < x < a + δ ; a ≠ x ∈ R }
= { x / 0 < |x-a| < δ ; x ∈ R }
- If , then N(a , ) ⊂ N(a , )
- N(a ,δ) ∩ N(b, δ) = ⟺ δ ≤ ; Where δ > 0
- f (x) = l
⇔∀ ε > 0, ∃ δ > 0 ∍
a – δ < x < a + δ, x ≠ a , x ∈ ⇒ l – ε < f (x) < l +
⇔∀ ε > 0, ∃ δ > 0 ∍
0 < |x-a| < δ , x ∈ ⟹ |f(x)-l| < ε
- f (x) = l ( Left hand limit )
⇔∀ ε > 0, ∃ δ > 0 ∍
x ∈ ( a – δ, a ) , x ∈ ⇒ < ε
⇔ ∀ ε > 0, ∃ δ > 0 ∍
a – δ < x < a, x ∈ ⇒ f (x) ∈ N( l , ε)
- f (x) = l ( Right hand limit )
⇔ ∀ ε > 0, ∃ δ > 0 ∍
x ∈ ( a , a + δ) , x ∈ < ε
⇔ ∀ ε > 0, ∃ δ > 0 ∍
a < x < a + δ, x ∈ ⇒ f (x) ∈ N( l , ε)
- Necessary and Sufficient Condition for Existence of f (x)
If f (x) and f (x) exists
and if f (x) = f (x) , then f (x) exists.
↦ In this case, f (x) = f (x) = f (x)
↦ And if f (x) exists, then its limit is unique ( one and only one ).
- If f (x) = k , then f (x) = k or k = k
- Working Rules of Limit :
If f (x) and g (x) exists, then
↦ [ f (x) ± g (x) ] = f (x) ± g (x)
↦ k f (x) = k · f (x) , k R
↦ [ f (x) · g (x) ] = f (x) · g (x)
↦ = ; Where g (x) ≠ 0
12. ; n ∈ R
13. = ; n ∈ R
14. = 1 or
15. OR
16. or
17. OR
- f is Continuous function on every a ∈ R ⇔ f (x) = f (a)
↦ Every Trigonometry Function and Inverse Trigonometry Function is Continuous on R.
- If f and g are Continuous functions on R ( or every a ∈ R ), then f ± g , f · g and ( Where, g (a) ≠ 0 ) are also Continuous functions.
- L’Hospital’s Rule :–
If we substitute x = a in both numerator (f ) and denominator (g) and we get the indeterminate form
( i.e. = or or ∞ × 0 or ∞ – ∞ or 00 or or ),
then taking derivative of both numerator and denominator separately w. r. t. x. And again substitute x = a and check its value. If we get again the indeterminate form, then again take derivative and so on. Continue the procedure till the required value is reached.