User:J D Bowen/For Julia

$$x_0\neq 0$$.

$$f'(x_0)=\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0} = \lim \frac{ x^2 \sin(1/x)- x_0^2 \sin(1/x_0)}{x-x_0} = 2x_0\sin(1/x_0)+x^2\cos(1/x_0)\ln|x_0|$$

$$f'(0)=\lim_{x\to 0} \frac{f(x)-0}{x-0} = \lim \frac{f(x)}{x} = \lim \frac{x^2 \sin(1/x)}{x} = \lim x\sin(1/x) = 0 \ $$.

$$s=t^p \ $$

$$\int_1^{x^p} \frac{dt}{t} = \int_1^x \frac{ds}{s} = \int_1^x \frac{pt^{p-1}dt}{t^p} = p \int_1^x \frac{t^{p-1}}{t^p}dt = p\int_1^x \frac{dt}{t} \ $$.

=1=

Define sup and inf from your text. $$\text{sup}(E) \ $$ being a real number is a real number that is a least upper bound, which implies $$E \ $$ has an upper bound, which implies $$E \ $$ is bounded above. similarly for inf.

=2=

Discussed in chat

=3=

$$\log(x)\leq Cx^p \iff y:=\log(x)/(x^p) \leq C \ $$.

Observe

$$y' = \frac{x^{p-1}-p\log(x)x^{p-1}}{x^{2p}} \ $$.

Setting $$y'=0 \ $$ yields

$$x^{p-1}-p\log(x)x^{p-1}= 0 \implies 1=p\log(x)\implies x=e^{1/p} \ $$.

Then $$f(e^{1/p})= \frac{1/p}{e} = 1/(pe) \ $$.

Observe that since $$C=1/(pe) \ $$ satisfies $$\frac{d}{dx} \log(x)/(x^p) (x=C) = 0 $$, this is either a maximum or a minimum. Further observe that since $$y'>0 \ $$ on $$(0,C) \ $$ and $$<0 \ $$ for $$x\in (C,\infty) \ $$, we must have $$C \ $$ as a maximum. Since $$\log(x)\leq C x^p \ $$ and since any $$C'<C \ $$ would fail to satisfy this in $$(f_-^{-1}(C'),f_+^{-1}(C')) \ $$, this constant satisfies the conditions of the problem.

=4=

Observe that $$(a-b)|(a-b) \ $$. Suppose that $$(a-b)|(a^n-b^n) \ $$. We have $$a^{n+1}-b^{n+1} = (a^n-b^n)(a+b) \ $$ and since $$(a-b)|(a^n-b^n) \ $$, we must have $$(a-b)|(a^n-b^n)(a+b)=a^{n+1}-b^{n+1} \ $$.

=5=

This clearly converges by the root test. Let $$L \ $$ be the limit. Then $$\lim_{n\to\infty} n^{1/n}=L \ $$. Take the logarithim of both sides to get $$\lim_{n\to\infty} 1/n = \log_n L \ $$. But $$\lim_{n\to\infty} 1/n= 0 \ $$, so $$\log_n L = 0 \implies L=n^0=1 \ $$.

=6=

By ratio test, the sequence converges because $$x_{n+1}/x_n = \frac{\prod_{j=1}^{n+1} (2j))}{\prod_{j=1}^{n+1} (2j+1)} \frac{\prod_{j=1}^n (2j+1)}{\prod_{j=1}^n (2j))} = \frac{2(n+1)}{2(n+1)+1)}  $$ which goes to 0 from below.

Further observe that $$x_n \in $$

=7=

Done in chat.

=8=

f,g cont if and only if $$\lim_{x\to a} f(x)=f(a), \lim_{x\to a}g(x)=g(a) \ $$ for all $$a \ $$. Since both are well defined at any value a and the quotient is as well, the quotient limit exists and we have

$$lim_{x\to a} f(x)/g(x) = \frac{\lim_{x\to a} f(x)}{\lim_{x\to a}g(x)} = f(a)/g(a) \ $$.

aNd so the quotient is continuous.

=9= For r = 0,


 * $$(1+x)^0 \ge 1+0x \, $$

is equivalent to 1 ≥ 1 which is true as required.

Now suppose the statement is true for r = k:


 * $$(1+x)^k \ge 1+kx. \, $$

Then it follows that



\begin{align} & {} \qquad (1+x)(1+x)^k \ge (1+x)(1+kx)\quad\text{(by hypothesis, since }(1+x)\ge 0) \\ & \iff (1+x)^{k+1} \ge 1+kx+x+kx^2, \\ & \iff (1+x)^{k+1} \ge 1+(k+1)x+kx^2. \end{align} $$

However, as 1 + (k + 1)x + kx2 ≥ 1 + (k + 1)x (since kx2 ≥ 0), it follows that (1 + x)k + 1 ≥ 1 + (k + 1)x, which means the statement is true for r = k + 1 as required.

By induction we conclude the statement is true for all r ≥ 0.

=10=

a) Suppose $$c\neq 0 \ $$. Then $$\frac{ax+b}{cx+d}\to \infty \ $$ as $$x\to -d/c \ $$ and so the function cannot be 1-1.  Suppose $$c=0 \ $$.  Then the function is just $$(a/d)x+(b/d) \ $$, a linear function which is necessarily one to one.

Observe that if $$y=\frac{ax+b}{d} \ $$, then $$x=\frac{d}{a}\left({y-\frac{b}{d}}\right) \ $$. This is the inverse function.

b) $$a=d=1, \ b=c=0 \ $$. Observe that then $$\frac{ax+b}{Cx+d}=x \ $$ which satisfies $$f^{-1}=f \ $$, and when any of the above inequalities fail to hold,

$$f^{-1}(x)=\frac{d}{a}\left({x-\frac{b}{d}}\right) \neq \frac{ax+b}{cx+d} \ $$.