Notes to a video lecture on http://www.unizor.com
Function Limit -
Standard Problems
Let's recall two definitions of a limit of function.
Definition 1
Value a is a limit of functionf(x) when its argument xconverges to real number r, if for ANY sequence of argument values {x_{n}} converging to r the sequence of function values {f(x_{n})} converges to a.
Symbolically:
∀{x_{n}}→r ⇒ {f(x_{n}}→a
Definition 2
For any positive ε there should be positive δ such that, if x is within
Symbolically:
∀ ε>0 ∃ δ>0:
|x−r| ≤ δ ⇒ |f(x)−a| ≤ ε
Solving the problems below, you can use any of these definitions to prove the existence of a limit and to find its concrete value.
Problem 1
Consider a function defined for all real arguments x and δ:
f(x) = [(x+δ)²−x²] / δ
Assume that variable x is fixed, while variable δ converges to 0.
Prove that this function has a limit for δ→0 and that this limit equals to 2x.
Solution
[(x+δ)²−x²] / δ =
= (x+δ−x)·(x+δ+x) / δ =
= δ·(2x+δ) / δ =
= 2x+δ
which converges to 2x as δ→0
Problem 2
Consider a function defined for all real arguments x and δ:
f(x) = [(x+δ)^{n}−x^{n}] / δ
Assume that variable x is fixed, while variable δ converges to 0.
Prove that this function has a limit for δ→0 and that this limit equals to n·x^{n−1}.
Solution
[(x+δ)^{n}−x^{n}] / δ =
= (x+δ−x)·
·[Σ_{j∈[0,n−1]}(x+δ)^{n−1−j}(x)^{j}] / δ =
= Σ_{j∈[0,n−1]}(x+δ)^{n−1−j}(x)^{j}
which converges to
= Σ_{j∈[0,n−1]}(x)^{n−1−j}(x)^{j} =
= n·x^{n−1}
Problem 3
For this problem you will need a theorem proven in theTrigonometry chapter (see lecture Geometry with Trigonometry - Lim sin(x)/x) that states that
sin(δ)/δ→1 if δ→0.
Consider a function defined for all real arguments x and δ:
f(x) = [sin(x+δ)−sin(x)] / δ
Assume that variable x is fixed, while variable δ converges to 0.
Prove that this function has a limit for δ→0 and that this limit equals to cos(x).
Solution
f(x) = [sin(x+δ)−sin(x)] / δ =
= (1/δ)[sin(x)cos(δ) +
+ cos(x)sin(δ)−sin(x)] =
= cos(x)·[sin(δ)/δ] −
− sin(x)·[1−cos(δ)]/δ
As we know, if δ→0,
sin(δ)/δ→1.
Therefore, the first component of the above expression is converging as follows
cos(x)·[sin(δ)/δ]→cos(x)
Considering
1−cos(δ) = 2sin²(δ/2),
we can write the following expressions:
[1−cos(δ)]/δ = 2sin²(δ/2)/δ =
= sin(δ/2)·[sin(δ/2)/(δ/2)]
The product of infinitesimal function sin(δ/2) (as δ→0) and a function sin(δ/2)/(δ/2)converging to 1 results in infinitesimal function.
Therefore,
sin(x)·[1−cos(δ)]/δ
is infinitesimal as δ→0 and our original function converges tocos(x)
Problem 4
Consider a function defined for all real arguments x and δ:
f(x) = [cos(x+δ)−cos(x)] / δ
Assume that variable x is fixed, while variable δ converges to 0.
Prove that this function has a limit for δ→0 and that this limit equals to −sin(x).
Solution
f(x) = [cos(x+δ)−cos(x)] / δ =
= (1/δ)[cos(x)cos(δ) −
− sin(x)sin(δ)−cos(x)] =
= −sin(x)·[sin(δ)/δ] −
− cos(x)·[1−cos(δ)]/δ
As we know, if δ→0,
sin(δ)/δ→1.
Therefore, the first component of the above expression is converging as follows
−sin(x)·[sin(δ)/δ]→−sin(x)
Considering
1−cos(δ) = 2sin²(δ/2),
we can write the following expressions:
[1−cos(δ)]/δ = 2sin²(δ/2)/δ =
= sin(δ/2)·[sin(δ/2)/(δ/2)]
The product of infinitesimal function sin(δ/2) (as δ→0) and a function sin(δ/2)/(δ/2)converging to 1 results in infinitesimal function.
Therefore,
cos(x)·[1−cos(δ)]/δ
is infinitesimal as δ→0 and our original function converges to−sin(x).
Problem 5
Consider a function defined for all non-negative arguments xand δ:
f(x) = [√(x+δ)−√x] / δ
Assume that variable x is fixed, while variable δ converges to 0.
Prove that this function has a limit for δ→0 and that this limit equals to 1/(2√x).
Solution
Multiply numerator and denominator of the original function by [√(x+δ)+√x].
The numerator will become[(x+δ)−x]=δ and it can be canceled with δ in denominator.
The result is
1 / [√(x+δ)+√x],
which converges to 1/(2√x) asδ→0.
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