MS 2001: Selected Limit Questions

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,

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

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MS 2001: Selected Limit Questions

Theorem

(ii)

Formula

Factor Theorem

Polynomial Division

(iii)

(iv)

(v)

(vi)

(vii)

Question 2 Summer 2010.

(a)

Closed Interval Method.

Using Inequalities

(b)

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J.P. McCarthy Maths

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MS 2001: Selected Limit Questions

Theorem

(ii)

Formula

Factor Theorem

Polynomial Division

(iii)

(iv)

(v)

(vi)

(vii)

Question 2 Summer 2010.

(a)

Closed Interval Method.

Using Inequalities

(b)

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J.P. McCarthy Maths

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