Worked examples · Sheet 1

Pointwise convergence of sequences of functions

Theory used: Sequence of functions Pointwise convergence

Exercise 1

Determine the pointwise limit on $R$ of the sequence of functions

f_{n} (x) = \frac{n x ^{2} + 1}{n} .

f_{n} (x) = x^{2} + \frac{1}{n}

n \to + \infty lim f_{n} (x) = x^{2} + n \to + \infty lim \frac{1}{n} = x^{2}

f (x) = x^{2} for every x \in R

The point $x$ is treated as a constant and the limit is computed in the variable $n$ alone. Dividing numerator and denominator by $n$ gives $f_{n} (x) = x^{2} + \frac{1}{n}$ , so the term $x^{2}$ does not depend on $n$ and the term $\frac{1}{n}$ tends to $0$ . Therefore, for each fixed $x \in R$ , $f_{n} (x)$ tends to $x^{2}$ , and the pointwise limit is the function $f (x) = x^{2}$ .

Exercise 2

Determine the pointwise limit on $[0, + \infty)$ of the sequence of functions

f_{n} (x) = \frac{x ^{n}}{1 + x ^{n}} .

0 \leq x < 1 : x^{n} \to 0, f_{n} (x) = \frac{x ^{n}}{1 + x ^{n}} \to \frac{0}{1 + 0} = 0

x = 1 : f_{n} (1) = \frac{1}{2} for every n

x > 1 : f_{n} (x) = \frac{1}{x ^{- n} + 1} \to \frac{1}{0 + 1} = 1

f (x) = ⎩ ⎨ ⎧ 0 \frac{1}{2} 1 if 0 \leq x < 1 if x = 1 if x > 1

y = \dfrac{x}{1 + x}

y = \dfrac{x^{4}}{1 + x^{4}}

y = \dfrac{x^{20}}{1 + x^{20}}

y = \frac{x ^{n}}{1 + x ^{n}}

for

n = 1, 4, 20

: all the graphs pass through

(1, \frac{1}{2})

and for large

n

they approach

0

for

x < 1

and

1

for

x > 1

The behavior of $x^{n}$ as $n \to + \infty$ depends on the base, so the limit is computed by distinguishing three cases. For $0 \leq x < 1$ the geometric sequence $x^{n}$ tends to $0$ , therefore the ratio tends to $0$ . For $x = 1$ every $f_{n}$ equals exactly $\frac{1}{2}$ , so the limit is $\frac{1}{2}$ . For $x > 1$ both the numerator and the denominator tend to $+ \infty$ ; dividing numerator and denominator by $x^{n}$ the ratio is rewritten as $\frac{1}{x ^{- n} + 1}$ , and since $x^{- n} \to 0$ the limit is $1$ . The pointwise limit is a function that is discontinuous at $x = 1$ , even though every $f_{n}$ is continuous.

Exercise 3

Determine the pointwise limit on $R$ of the sequence of functions

f_{n} (x) = (1 + \frac{x}{n})^{n} .

f_{n} (x) = e^{n l n (1 + \frac{x}{n})} for n > ∣ x ∣

n ln (1 + \frac{x}{n}) = x \cdot \frac{ln ( 1 + \frac{x}{n} )}{\frac{x}{n}} n \to + \infty x \cdot 1 = x (x \neq = 0)

n \to + \infty lim f_{n} (x) = e^{x} for every x \in R

For each fixed $x$ and for $n > ∣ x ∣$ the base $1 + \frac{x}{n}$ is positive, so the power is rewritten in exponential form as $f_{n} (x) = e^{n l n (1 + \frac{x}{n})}$ . The exponent reduces to the standard limit of the logarithm: setting $t = \frac{x}{n}$ , as $n \to + \infty$ we have $t \to 0$ , so $\frac{ln ( 1 + t )}{t} \to 1$ and the exponent tends to $x$ . For $x = 0$ every $f_{n}$ equals $1 = e^{0}$ , so the formula remains valid. Since the exponential is a continuous function, the limit of the exponent gives $f_{n} (x) \to e^{x}$ for every $x \in R$ .