# Math Help - Revision

1. ## Revision

1. Show that f(x) is an even function where: $f(x) = ln(1+e^x) - x/2$

2. Evaluate lim x--> 5 [sqrt(x) - sqrt(5)]/[x-5]

3. Prove, by induction, that a^n - b^n is divisible by a-b for all positive integer values of n.

ARGH i'm so stupid. Could some1 plz help me with these questions???

2. 1. $f(-x)=\ln(1+e^{-x})+\frac{x}{2}=\ln\left(1+\frac{1}{e^x}\right)+\f rac{x}{2}=$

$=\ln\frac{e^x+1}{e^x}+\frac{x}{2}=\ln(1+e^x)-\ln e^x+\frac{x}{2}=$

$=\ln(1+e^x)-x+\frac{x}{2}=\ln(1+e^x)-\frac{x}{2}=f(x)$

2. $\lim_{x\to 5}\frac{\sqrt{x}-\sqrt{5}}{x-5}=\lim_{x\to 5}\frac{\sqrt{x}-\sqrt{5}}{(\sqrt{x}-\sqrt{5})(\sqrt{x}+\sqrt{5})}=$

$=\lim_{x\to 5}\frac{1}{\sqrt{x}+\sqrt{5}}=\frac{1}{2\sqrt{5}}$

3. For n=1: $a-b\vdots a-b$

Suppose that $a^n-b^n\vdots a-b$

$a^{n+1}-b^{n+1}=a^{n+1}-ab^n+ab^n-b^{n+1}=$

$=a(a^n-b^n)+b^n(a-b)$

$a^n-b^n\vdots a-b\Rightarrow a(a^n-b^n)\vdots a-b$ (1)

$b^n(a-b)\vdots a-b$ (2)

From (1) and (2) $a^{n+1}-b^{n+1}\vdots a-b$

3. Originally Posted by xwrathbringerx
1. Show that f(x) is an even function where: $f(x) = ln(1+e^x) - x/2$

2. Evaluate lim x--> 5 [sqrt(x) - sqrt(5)]/[x-5]

3. Prove, by induction, that a^n - b^n is divisible by a-b for all positive integer values of n.

ARGH i'm so stupid. Could some1 plz help me with these questions???
1. $f(-x) = \ln (1 + e^{-x}) - \frac{(-x)}{2} = \ln \left( 1 + \frac{1}{e^x} \right) + \frac{x}{2} = \ln \left( \frac{e^x + 1}{e^x} \right) - \frac{x}{2}$

$= \ln (e^x + 1) - \ln e^x + \frac{x}{2} = \ln (e^x + 1) - x + \frac{x}{2} = f(x)$.

2. Note that $x - 5 = (\sqrt{x} - \sqrt{5}) (\sqrt{x} + \sqrt{5})$.

3. Show true for n = 1. Assume true for n = k. Show that it follows from the assumption that it's true for n = k+1.

4. Originally Posted by xwrathbringerx
1. Show that f(x) is an even function where: $f(x) = ln(1+e^x) - x/2$

2. Evaluate lim x--> 5 [sqrt(x) - sqrt(5)]/[x-5]

3. Prove, by induction, that a^n - b^n is divisible by a-b for all positive integer values of n.

ARGH i'm so stupid. Could some1 plz help me with these questions???
Note that $x - 5 = (\sqrt{x} + \sqrt{5})(\sqrt{x} - \sqrt{5})$.

So $\frac{\sqrt{x} - \sqrt{5}}{x - 5} = \frac{\sqrt{x} - \sqrt{5}}{(\sqrt{x} + \sqrt{5})(\sqrt{x} - \sqrt{5})} = \frac{1}{\sqrt{x} + \sqrt{5}}$.

Therefore $\lim_{x \to 5}\frac{\sqrt{x} - \sqrt{5}}{x - 5} = \lim_{x \to 5}\frac{1}{\sqrt{x} + \sqrt{5}}$

$= \frac{1}{\sqrt{5} + \sqrt{5}}$

$= \frac{1}{2\sqrt{5}}$

$= \frac{\sqrt{5}}{10}$.