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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