Tag: Derivatives

Show that the limit and derivative cannot be interchanged when f_n(x) = (sin (nx)) / n

by RoRi

March 31, 2016

For each positive integer $n$ and and all real $x$ define

$f_n (x) = \frac{\sin (nx)}{n}, \qquad f(x) = \lim_{n \to \infty} f_n(x).$

Prove that

$\lim_{n \to \infty} f_n'(0) \neq f'(0).$

Proof. First, we have

$f_n (x) = \frac{\sin (nx)}{n} \quad \implies \quad f_n'(x) = \cos (nx) \quad \implies \quad f_n'(0) = 1$

for all $n$ . Hence,

$\lim_{n \to \infty} f_n'(0) = 1.$

On the other hand,

$f(x) = \lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{\sin (nx)}{n} = 0 \implies f'(0) = 0.$

Hence,

$\lim_{n \to \infty} f_n'(0) \neq f'(0). \qquad \blacksquare$

Prove some properties of the function e^-1/x²

by RoRi

January 13, 2016

Consider the function

$f(x) = e^{-\frac{1}{x^2}} \qquad \text{when} \quad x \neq 0$

and $f(0) = 0$ .

Prove that for every positive number $m$ we have
$\lim_{x \to 0} \frac{f(x)}{x^m} = 0.$
Prove that if $x \neq 0$ then
$f^{(n)}(x) = f(x) P \left( \frac{1}{x} \right)$

where $P(t)$ is a polynomial in $t$ .
Prove that
$f^{(n)}(0) = 0 \qquad \text{for all } n \geq 1.$

Proof. (A specific case of this general theorem is actually the first problem of this section, here. Maybe it’s worth taking a look since this proof is just generalizing that particular case.) We make the substitution $t = \frac{1}{x^2}$ , so that $t \to +\infty$ as $x \to 0$ and we have
$\begin{align*} \lim_{x \to 0} \frac{e^{-\frac{1}{x^2}}}{x^m} &= \lim_{t \to +\infty} \frac{t^{\frac{m}{2}}}{e^t} \\[9pt] &= 0 \end{align*}$

by Theorem 7.11 (page 301 of Apostol) since $m > 0$ implies $\frac{m}{2} > 0$ as well $. \qquad \blacksquare$
Proof. The proof is by induction on $n$ . In the case $n = 1$ we have
$\begin{align*} f'(x) &= \left( \frac{2}{x^3} \right) e^{-\frac{1}{x^2}} \\[9pt] &= 2 \left( \frac{1}{x} \right)^3 e^{-\frac{1}{x^2}} \\[9pt] &= P \left(\frac{1}{x} \right) e^{-\frac{1}{x^2}}. \end{align*}$

So, indeed the formula is valid in the case $n = 1$ . Assume then that the formula holds for some positive integer $k$ . We want to show this implies the formula holds for the case $k + 1$ .

$\begin{align*} f^{(k+1)}(x) = \left( f^{(k)}(x) \right)' &= \left( P\left( \frac{1}{x} \right) e^{-\frac{1}{x^2}} \right)' \\[9pt] &= P' \left( \frac{1}{x} \right) e^{-\frac{1}{x^2}} + \frac{2}{x^3} P \left( \frac{1}{x} \right) e^{-\frac{1}{x^2}} \\[9pt] &= \left( P' \left( \frac{1}{x} \right) + \frac{2}{x^3} P \left( \frac{1}{x} \right) \right)e^{-\frac{1}{x^2}}. \end{align*}$

But then the leading term

$P' \left( \frac{1}{x} \right) + 2 \left( \frac{1}{x} \right)^3 P \left( \frac{1}{x} \right)$

is still a polynomial in $\frac{1}{x}$ since the derivative of a polynomial in $\frac{1}{x}$ is still a polynomial in $\frac{1}{x}$ , and so is the sum of two polynomials in $\frac{1}{x}$ . Therefore, we have that the formula holds for the case $k+1$ ; hence, it holds for all positive integers $n. \qquad \blacksquare$
Proof. The proof is by induction on $n$ . If $n = 1$ then we use the limit definition of the derivative to compute the derivative at 0,
$\begin{align*} f'(0) &= \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \\[9pt] &= \lim_{h \to 0} \frac{f(h)}{h} &(f(0) = 0 \text{ by def of } f)\\[9pt] &= 0 & \text{by part (a)}. \end{align*}$

So, indeed $f'(0) = 0$ and the statement is true for the case $n = 1$ . Assume then that $f^{(k)} (0) = 0$ for some positive integer $k$ . Then, we use the limit definition of the derivative again to compute the derivative $f^{(k+1)}(0)$ ,

$\begin{align*} f^{(k+1)}(0) &= \lim_{h \to 0} \frac{f^{(k)}(0+h) - f^{(k)}(0)}{h} \\[9pt] &= \lim_{h \to 0} \frac{f^{(k)}(h)}{h} &(f^{(k)}(0) = 0 \text{ by Ind. Hyp.}) \\[9pt] &= \lim_{h \to 0} f(h) P \left( \frac{1}{h} \right) \cdot \frac{1}{h} &(\text{part (b)}) \\[9pt] &= \lim_{h \to 0} P \left( \frac{1}{h} \right) e^{-\frac{1}{h^2}}. \end{align*}$

This follow since $\frac{1}{h} \cdot P \left( \frac{1}{h} \right)$ is still a polynomial in $\frac{1}{h}$ , and by the definition of $f(x)$ for $x \neq 0$ . But then, by part (a) we know

$\lim_{h \to 0} \frac{f(h)}{h^m} = 0 \qquad \text{for all } m \in \mathbb{Z}^+.$

Therefore,

$f^{(k+1)}(0) = \lim_{h \to 0} P \left( \frac{1}{h} \right) e^{-\frac{1}{h^2}} = 0.$

Thus, the formula holds for the case $k+1$ , and hence, for all positive integers $n. \qquad \blacksquare$

Find the limit of the given function

by RoRi

December 22, 2015

Find the value of the following limit.

$\lim_{x \to 1^+} \frac{x^x - x}{1 - x + \log x}.$

We apply L’Hopital’s rule (twice) to compute the limit (Noting that, $(x^x - x)' = x^x + x^x \log x -1$ ),

$\begin{align*} \lim_{x \to 1^+} \frac{x^x - x}{1 - x + \log x} &= \lim_{x \to 1^+}\frac{x^x + x^x \log x - 1}{\frac{1}{x} -1} \\[9pt] &= \lim_{x \to 1^+} \frac{x^{x+1} + x^{x+1} \log x - x}{1 - x} \\[9pt] &= \lim_{x \to 1^+} \frac{x^{x+1} \left( \log x + \frac{x+1}{x} \right) + x^{x+1} \left( \log x + \frac{x+1}{x} \right) + x^x - 1}{-1} \\[9pt] &= -2. \end{align*}$

Since this limit exists, the application of L’Hopital’s rule was justified.

Find the limit of the given function

by RoRi

December 22, 2015

Find the value of the following limit.

$\lim_{x \to a^+} \frac{ \sqrt{x} - \sqrt{a} + \sqrt{x-a}}{\sqrt{x^2-a^2}}.$

We compute as follows,

$\begin{align*} \lim_{x \to a^+} \frac{\sqrt{x} - \sqrt{a} + \sqrt{x-a}}{\sqrt{x^2-a^2}} &= \lim_{x \to a^+} \left( \frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 - a^2}} + \frac{\sqrt{x-a}}{\sqrt{(x+a)(x-a)}} \right) \\[9pt] &= \lim_{x \to a^+} \left( \frac{\sqrt{x} - \sqrt{a}}{\sqrt{x^2 - a^2}} \right) + \frac{1}{2\sqrt{a}} \\[9pt] &= \frac{1}{2 \sqrt{a}} + \lim_{x \to a^+} \frac{ \frac{1}{2 \sqrt{x}}}{\frac{x}{\sqrt{x^2-a^2}}} &(\text{L'Hopital})\\[9pt] &= \frac{1}{2\sqrt{a}} + \lim_{x \to a^+} \frac{\sqrt{x^2-a^2}}{2x^{\frac{3}{2}}} \\[9pt] &= \frac{1}{2\sqrt{a}} + 0 = \frac{1}{2\sqrt{a}}. \end{align*}$

Since this limit exists, the application of L’Hopital’s rule was justified.

Find the limit of the given function

by RoRi

December 22, 2015

Find the value of the following limit.

$\lim_{x \to 0^+} \frac{x - \sin x}{(x \sin x)^{\frac{3}{2}}}.$

First, since $\lim_{x \to 0} \frac{\sin x}{x} = 1$ we can multiply by this inside the limit without changing the value,

$\begin{align*} \lim_{x \to 0^+} \frac{x - \sin x}{(x \sin x)^{\frac{3}{2}}} &= \lim_{x \to 0^+} \left( \frac{x - \sin x}{(x \sin x)^{\frac{3}{2}}} \cdot \left( \frac{\sin x}{x} \right)^{\frac{3}{2}} \right) \\[9pt] &= \lim_{x \to 0^+} \frac{x - \sin x}{x^3}. \end{align*}$

Now we can apply L’Hopital’s rule to compute the limit,

$\begin{align*} \lim_{x \to 0^+} \frac{x - \sin x}{x^3} &= \frac{ 1 - \cos x}{3x^2} \\[9pt] &= \lim_{x \to 0^+} \frac{ \sin x}{6x} \\[9pt] &= \lim_{x \to 0^+} \frac{\cos x}{6} \\[9pt] &= \frac{1}{6}. \end{align*}$

Since this limit exists, the application of L’Hopital’s rule was justified.

Find the limit of the given function

by RoRi

December 22, 2015

Find the value of the following limit.

$\lim_{x \to 0} \frac{\log(\cos (ax))}{\log (\cos (bx))}.$

Applying L’Hopital’s rule we have,

$\begin{align*} \lim_{x \to 0} \frac{\log(\cos (ax))}{\log(\cos (bx))} &= \lim_{x \to 0} \left( \frac{a}{b} \cdot \frac{\cos (bx)}{\cos (ax)} \cdot \frac{\sin (ax)}{\sin (bx)} \right) \\[9pt] &= \frac{a}{b} \lim_{x \to 0} \frac{a \cos (ax)}{b \cos (bx)} \\[9pt] &= \left(\frac{a}{b} \right)^2. \end{align*}$

(We used that $\lim_{x \to 0} \frac{\sin (ax)}{\sin (bx)} = \frac{a}{b}$ from this exercise, Section 7.11, Exercise #1). Since this limit exists, the application of L’Hopital’s rule was justified.

Find the limit of the given function

by RoRi

December 22, 2015

Find the value of the following limit.

$\lim_{x \to 0} \frac{(2-x)e^x - x - 2}{x^3}.$

We apply L’Hopital’s rule three times,

$\begin{align*} \lim_{x \to 0} \frac{(2-x)e^x - x - 2}{x^3} &= \lim_{x \to 0} \frac{(2-x)e^x - e^x - 1}{3x^2} \\[9pt] &= \lim_{x \to 0} \frac{(2-x)e^x - 2e^x}{6x} \\[9pt] &= \lim_{x \to 0} \frac{(2-x)e^3 - 3e^x}{6} \\[9pt] &= -\frac{1}{6}. \end{align*}$

Since this final limit exists, the chain of applications of L’Hopital’s was justified.

Find the limit of the given function

by RoRi

December 22, 2015

Find the value of the following limit.

$\lim_{x \to 0} \frac{\sinh x - \sin x}{x^3}.$

We will apply L’Hopital’s rule three times,

$\begin{align*} \lim_{x \to 0} \frac{\sinh x - \sin x}{x^3} &= \lim_{x \to 0} \frac{\cosh x - \cos x}{3x^2} \\[9pt] &= \lim_{x \to 0} \frac{\sinh x + \sin x}{6x} \\[9pt] &= \lim_{x \to 0} \frac{\cosh x + \cos x}{6} \\[9pt] &= \frac{1}{3}. \end{align*}$