Questions 1 (ii) — (vii) from here.

With all of these questions the first thing we want to do is plug in the value; i.e. if we are evaluating $\lim_{x\rightarrow a}f(x)$ we should first try $f(a)$. If we find that $f(a)$ is undefined (e.g. division by zero), we will usually have to ‘factor out the bad stuff’. The reason this works is because of the following theorem:

### Theorem

Suppose that $f(x)=g(x)$ except perhaps at $x=a$. Then

$\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}g(x)$.

Take for example

$\lim_{x\rightarrow 0}\frac{x^2}{x}$.

Now if $f(x)=x^2/x$, then $f(0)=0/0$ — which is undefined. However we can cancel an $x$ above and below… well what we really do is as follows:

$\frac{x^2}{x}=x\times\underbrace{\frac{x}{x}}_{=1}=x$.

However this is only true in the case where $x\neq 0$; i.e. we have that $x^2/x=x$ for all $x\neq 0$ — and use the above theorem to say that

$\lim_{x\rightarrow 0}\frac{x^2}{x}=\lim_{x\rightarrow 0}x=0$.

## (ii)

$\lim_{t\rightarrow-2}\frac{t^3+8}{t^2-4}$.

First off plugging in $-2$ gives $0/0$ — undefined. What we want to do is factorise above and below and hopefully there will be a $(t+2)=(t-(-2))$ term which we can cancel above and below (for $t\neq -2$), and use the above theorem. First off the denominator is simply a difference of two squares:

$t^2-4=t^2-2^2=(t-2)(t+2)$.

I can think of three ways to factorise the numerator.

#### Formula

The numerator has the form $a^3+b^3$ which I know has factors via a formula. I never learnt this formula ‘off’ but instead have a quick derivation (using the Binomial Theorem I suppose):

$(a+b^3)=a^3+3a^2b+3ab^2+b^3$

$\Rightarrow a^3+b^3=(a+b)^3-3a^2b-3ab^2$

$=(a+b)^2-3ab(a+b)=(a+b)[(a+b)^2-3ab]$

$=(a+b)(a^2+2ab+b^2-3ab)$

$\Rightarrow a^3+b^3=(a+b)(a^2-ab+b^2)$.

Hence I have

$t^3+2^3=(t+2)(t^2-2t+4)$.

#### Factor Theorem

The factor theorem states that:

A complex number $k$ is a root of a polynomial $p(x)$ if and only if $(x-k)$ is a factor of $p(x)$.

i.e. if $p(k)=0$, then $p(x)=(x-k)q(x)$ for some polynomial $q(x)$ (and vice versa).

When I plugged in $-2$ into $t^3+8$ I got zero — hence $(t-(-2))=(t+2)$ is a factor of $t^3+8$. This means I can perform polynomial long division to get

$t^3+0t^2+0t+4\div t+2=t^2-2t+4$;

that is

$t^3+8=(t+2)(t^2-2t+4)$

as before.

#### Polynomial Division

You can just straight away do polynomial long division to get

$t^3+0t^2+0t+8\div t^2+0t-4=t+\frac{4t+8}{t^2-4}$

$\Rightarrow \frac{t^3+8}{t^2-4}=t+4\frac{t+2}{(t-2)(t+2)}$

$=t+\frac{4}{t-2}$,

for $t\neq -2$.

Taking one of the first two factorisations leads to:

$\frac{t^3+8}{t^2-4}=\frac{(t+2)(t^2-2t+4)}{(t+2)(t-2)}=\frac{t^2-2t+4}{t-2}$,

for $t\neq-2$.

Using this or the $t+4/(t-2)$ will allow us then to plug in $t=-2$ to get

$\lim_{t\rightarrow -2}\frac{t^3+8}{t^2-4}=-3$.

## (iii)

Plugging in $r=3$ yields division by zero — undefined.

$\frac{r^2-9}{r^2-5r+6}=\frac{(r-3)(r+3)}{r^2-3r-2r+6}$

$=\frac{(r-3)(r+3)}{r(r-3)-2(r-3)}=\frac{(r-3)(r+3)}{(r-3)(r-2)}$

$=\frac{r+3}{r-2}$,

for $r\neq 3$. Hence:

$\lim_{r\rightarrow 3}\frac{r^2-9}{r^2-5r+6}=\lim_{r\rightarrow3}\frac{r+3}{r-2}=6$.

## (iv)

Plugging in $r=-1$ gives $-3/4$. Done and dusted!

## (v)

Plugging in $x=0$ gives $\frac{\sqrt{3}}{-3}=-\frac{1}{\sqrt{3}}$ — done.

## (vi)

Plugging in $x=3$ gives division by zero. These type need a manipulation called ‘multiplying by the conjugate’ (which is just multiplying by $1$ — which changes nothing of course):

$\frac{\sqrt{2x+3}-x}{x-3}\times\frac{\sqrt{2x+3}+x}{\sqrt{2x+3}+x}$

$=\frac{(\sqrt{2x+3}-x)(\sqrt{2x+3}+x)}{(x-3)(\sqrt{2x+3}+x)}$.

The numerator here is a difference of two squares: $(a-b)(a+b)=a^2-b^2$ so we have:

$\frac{2x+3-x^2}{(x-3)(\sqrt{2x+3}+x)}$.

Hopefully the top will factorise with an $(x-3)$ term that we could cancel out:

$-x^2+2x+3=-x^2+3x-x+3=-x(x-3)-1(x-3)=(x-3)(-x-1)$.

Thence we have

$\frac{(x-3)(-x-1)}{(x-3)(\sqrt{2x+3}+x)}=\frac{-x-1}{\sqrt{2x+3}+x}$,

for $x\neq3$. Therefore

$\lim_{x\rightarrow 3}\frac{\sqrt{2x+3}-x}{x-3}=\lim_{x\rightarrow 3}\frac{-x-1}{\sqrt{2x+3}+x}=-\frac{2}{3}$.

## (vii)

This one is not so straightforward and beyond exam standard. I can’t see a simple solution I will simply present a solution.

At $h=1$, the expression in undefined. Let $y=\sqrt{h}$ and consider

$\frac{h^5-3h+2}{\sqrt{h}-h}=\frac{y^{10}-3y^2+2}{y-y^2}$,

$=-\frac{y^{10}-3y^2+2}{y(y-1)}$.

By inspection $y=1$ is a root of $y^{10}-3y^2+2$ so hence $(y-1)$ is a factor. Polynomial long division yields:

$(y^{10}-3y^2+2)\div (y-1)$

$=y^9+y^8+y^7+y^6+y^5+y^4+y^3+y^2-2y-2$.

Therefore, after a cancellation of the $(y-1)$ terms, and division by a $y$:

$-\frac{y^{10}-3y^2+2}{y(y-1)}=-\left(y^8+y^7+y^6+y^5+y^4+y^3+y^2+y-2-\frac{2}{y}\right)$,

$=-\left(\sqrt{h}^8+\sqrt{h}^7+\sqrt{h}^6+\sqrt{h}^5+\sqrt{h}^4+\sqrt{h}^3+\sqrt{h}^2+\sqrt{h}-2-\frac{2}{\sqrt{h}}\right)$

Now plugging in $h=1$ gives $-4$.

## Question 2 Summer 2010.

### (a)

We want to find positive integers (whole numbers) $M$ and $N$ such that:

$M\leq\left|\frac{9x-2}{7x+4}\right|\leq N$.

There are two main methods of doing this.

#### Closed Interval Method.

Suppose that a function $f:[a,b]\rightarrow\mathbb{R}$ is continuous. Then $f$ attains its absolute maximum and minimum on $[a,b]$ and further can only be attained at criticial points — i.e. the endpoints $a,b$, points where the derivative is zero ($f'=0$), or points where the function is not differentiable ($f'$ does not exist).

In this case we want to look at $f(x)=(9x-2)/(7x+4)$. We know that the numerator and denominator are continuous as polynomials and their quotient is continuous for $7x+4\neq 0\Leftrightarrow x\neq-4/7$ — which is certainly the case for $x\in[1,4]$. Hence we can use the closed interval method to find the absolute maximum/ minimum of $f(x)$.

First we find the critical points. As the endpoints, we have $x=1,4$ are critical. Next we find points where $f'(x)=0$. By the quotient rule:

$f'(x)=\frac{(7x+4)9-(9x-2)7}{(7x+4)^2}=\frac{50}{(7x+4)^2}$.

Now a fraction is equal to zero if and only if the numerator is zero and as $50\neq 0$ (!), there are no points where $f'=0$.

Finally, by the quotient rule, $f(x)$ is differentiable for all $7x+4\neq0$ which is satisfied for all $x\in[1,4]$ so the function is differentiable on $[1,4]$.

Hence the only critical points are $1$ and $4$ so we test the function at these points:

$f(1)=\frac{9(1)-2}{7(1)+4}=\frac{7}{11}$.

$f(4)=\frac{9(4)-2}{7(4)+4}=\frac{34}{32}=\frac{17}{16}$.

Now these are the absolute maximum and minimum of $f(x)$ on $[1,4]$ so we can safely conclude that

$0\leq|f(x)|\leq 2$

on $[1,4]$ as required.

#### Using Inequalities

A quick manipulation using the properties of the absolute value function:

$\left|\frac{9x-2}{7x+4}\right|=\frac{|9x-2|}{|7x+4|}$.

This problem requires the following fact:

For $a,b,c,d\in\mathbb{R}$, if $0\leq a< b$ and $0\leq c< d$ then we may divide the smaller by the larger and the larger by the smaller to preserve the inequality,  i.e.

$\frac{a}{d}\leq \frac{b}{c}$.

So to find an upper bound ($N$), I want to find an upper bound for $|9x-2|$ and a lower bound for $|7x+4|$.

By the triangle inequality, and using properties of the absolute value function:

$|9x-2|\leq|9x|+|-2|=9|x|+2$.

Clearly this is maximised for large $x$ so we have (i.e. $x=4$ is the largest $x$ we are concerned with in $[1,4]$):

$|9x-2|\leq 38$.

By the reverse triangle inequality:

$|7x+4|\geq |7x|-|4|=7|x|-4\geq 3$,

for similar reasons.

Putting these two bounds together, using the above fact:

$\left|\frac{9x-2}{7x+4}\right|\leq\frac{38}{3}\leq 13$.

Now we do the opposite. First a lower bound for $|9x-2|$:

$|9x-2|\geq |9x|-|2|=9|x|-2\geq 7$.

A upper bound for $|7x+4|$:

$|7x+4|\leq |7x|+|4|=7|x|+4\leq 32$.

So we have:

$\left|\frac{9x-2}{7x+4}\right|\geq \frac{7}{32}\geq 0$.

So, by this method, all and all:

$0\leq \left|\frac{9x-2}{7x+4}\right|\leq 13$.

(Note that the closed interval method gives exactly sharp bounds (well until we round to the smaller/ larger integer)).

### (b)

Now let $g(x)=(x^2-3x)/(7x+4)$ and $L=-1/9$. We want to show that for all $\varepsilon>0$, there exists a $\delta>0$ such that if $0<|x-2|<\delta$, then $|g(x)-L|<\varepsilon$.

How we start this is by looking at the $|g(x)-L|$ term.

$g(x)-L=\frac{x^2-3x}{7x+4}-\left(-\frac{1}{9}\right)$

$=\frac{9}{9}\cdot\frac{x^2-3x}{7x+4}+\frac{1}{9}\cdot\frac{7x+4}{7x+4}$

$=\frac{9(x^2-3x)+7x+4}{9(7x+4)}=\frac{9x^2-20x+4}{9(7x+4)}$.

We would hope at this stage to see a $(x-2)$ term and a $(9x-2)/(7x+4)$ term. Sure enough if we factorise the numerator we see $9x^2-20x+4=(9x-2)(x-2)$ so we write:

$|g(x)-L|=\left|\frac{(9x-2)(x-2)}{9(7x+4)}\right|$,

using the properties of the absolute value function:

$|g(x)-L|=\frac{1}{9}\left|\frac{9x-2}{7x+4}\right||x-2|$.

This is exactly where we want to be now because we can use the upper bound above ($N$ — be it $2$ or $13$ depending on which method we used) to say that

$|g(x)-L|<\frac{1}{9}N|x-2|$.

Hence now if we make $|x-2|$ small so that

$\frac{1}{9}N|x-2|\sim \varepsilon$,

$\Leftrightarrow |x-2|\sim \frac{9\varepsilon}{N}$,

i.e. $|x-2|<9\varepsilon/N=:\delta$, then we are done.

Now we write:

Let $\varepsilon>0$. Choose $\delta=9\varepsilon/N$ ($N$ could be $2$ or $13$ or whatever upper bound you found in part (a)). Now let $0<|x-2|<\delta$. Then

$|g(x)-L|=\frac{1}{9}|f(x)||x-2|<\frac{1}{9}N\frac{9\varepsilon}{N}=\varepsilon$.

That is,

$\lim_{x\rightarrow 2}\frac{x^2-3x}{7x+4}=-\frac{1}{9}$ $\bullet$