TC10–TC12 - Limits, asymptotes and cusps | OCR A-Level Further Mathematics B (MEI)

Use CAS limits and nearby behaviour to distinguish holes from vertical asymptotes, find horizontal and oblique asymptotes, and identify asymptotic approach to another curve. Recognise MEI cusps by the two branches and their common limiting tangent.

A missing input need not give an infinite branch

Can a curve cross its horizontal asymptote infinitely many times? The graph of $y=1+\sin x/x$ , $x>0$ , does exactly that. We will explain why this does not contradict approaching the line.

First investigate finite inputs. A denominator becoming zero tells you to examine a limit; it does not by itself decide whether a vertical asymptote exists.

You already know

Keep an original equation's domain after cancellation. CAS can take a limit at a specified input or at infinity, but a left-hand limit and a right-hand limit may differ. Tangent gradients depend on the curve representation and may need a limiting calculation when a direct quotient is inconclusive.

Keep the value at a point separate from the behaviour of nearby points. A limit can exist even when the original formula supplies no value there.

Finite-input behaviour

A finite limit

\lim_{x\to a}f(x)=L

describes values approaching $L$ , even if $f(a)$ is undefined. If $f(a)$ is missing, the graph may have a hole at $(a,L)$ .

If at least one one-sided limit tends to $+\infty$ or $-\infty$ , $x=a$ is a vertical asymptote. A two-sided finite limit requires the left and right limits to agree.

Not in the formula booklet: learn it.

For the family $f_a(x)=(x^2+a)/(x-1)$ , $x\ne1$ , compare the CAS outputs below. Specify the side explicitly rather than relying on an unspecified default.

Parameter	Limit as $x\to1^-$	Limit as $x\to1^+$
$a=1$	$-\infty$	$+\infty$
$a=-1$	$2$	$2$

Worked example 1: the non-cancelling member

Question. Describe the behaviour of $f_1(x)=(x^2+1)/(x-1)$ near $x=1$ .

Plan. Check the numerator's limiting sign, then approach through negative and positive denominator values.

\begin{aligned} x^2+1&\longrightarrow2>0&&\text{numerator near }1\\ x\to1^-&\Longrightarrow f_1(x)\to-\infty &&\text{small negative denominator}\\ x\to1^+&\Longrightarrow f_1(x)\to+\infty &&\text{small positive denominator} \end{aligned}

Why this step? The signs describe two different branches. Writing only “infinity” would lose which way each one goes.

Answer. $x=1$ is a vertical asymptote; the left branch tends downwards without bound and the right branch upwards without bound.

Check. $f_1(0.99)=-198.01$ , while $f_1(1.01)=202.01$ , consistent with the limits.

Change only the numerator parameter from $1$ to $-1$ .

Worked example 2: cancellation leaves a hole

Question. Describe $f_{-1}(x)=(x^2-1)/(x-1)$ near $x=1$ .

Plan. Simplify where permitted and take the limit without filling the excluded input.

\begin{aligned} f_{-1}(x)&=\frac{(x-1)(x+1)}{x-1}=x+1,\quad x\ne1 &&\text{retain the original domain}\\ \lim_{x\to1^-}f_{-1}(x)&=2,\qquad \lim_{x\to1^+}f_{-1}(x)=2 &&\text{matching finite limits} \end{aligned}

Why this step? A limit describes nearby values; it does not automatically supply the value at the excluded input.

Answer. A hole at $(1,2)$ , with no vertical asymptote there.

Check. The nearby values are $1.99$ and $2.01$ . Neither tends to an infinite magnitude.

A plot can hide a hole or appear to join across an asymptote. Exclude the original input, use separate branches and verify the proposed behaviour with limits.

Watch out

Wrong: “denominator zero, therefore vertical asymptote”. Right: simplify with conditions and inspect one-sided limits; a cancelled factor can instead leave a hole.

Change only the numerator parameter again.

Check the difference from a proposed line

A curve approaches the line $y=mx+b$ if its difference from that line tends to zero along the relevant infinite branch. Check $x\to+\infty$ and $x\to-\infty$ separately; one branch's answer need not describe the other.

Horizontal and oblique asymptotes

\lim_{x\to\pm\infty}f(x)=b,

then $y=b$ is a horizontal asymptote on the corresponding branch. For a proposed oblique line with $m\ne0$ , check

\lim_{x\to\pm\infty}\bigl(f(x)-(mx+b)\bigr)=0

One way to find its coefficients, when the limits exist finitely, is

m=\lim_{x\to\pm\infty}\frac{f(x)}x,\qquad b=\lim_{x\to\pm\infty}\bigl(f(x)-mx\bigr)

Polynomial division can identify the same line directly.

Not in the formula booklet: learn it.

For $g_a(x)=(ax^2+2x+3)/(x+1)$ , $x\ne-1$ , CAS division gives

g_a(x)=ax+(2-a)+\frac{a+1}{x+1}

The residual term makes both the limiting line and the side of approach visible.

Worked example 3: a horizontal limiting line

Question. Find the infinite-branch asymptote of $g_0(x)=(2x+3)/(x+1)$ .

Plan. Use the divided form and take limits of its reciprocal remainder.

\begin{aligned} g_0(x)&=2+\frac1{x+1}&&\text{CAS division}\\ g_0(x)-2&=\frac1{x+1}\longrightarrow0 &&\text{as }x\to\pm\infty \end{aligned}

Why this step? The constant part is the line approached; the remainder proves the approach rather than just suggesting it from a graph.

Answer. Horizontal asymptote $y=2$ at both infinite ends.

Check. The curve is above the line for $x>-1$ and below it for $x<-1$ , matching the remainder's sign.

Change only the quadratic coefficient from $a=0$ to $a=1$ .

Worked example 4: an oblique limiting line

Question. Find the infinite-branch asymptote of $g_1(x)=(x^2+2x+3)/(x+1)$ .

Plan. Extract the linear part, then verify that subtracting it leaves a vanishing residual.

\begin{aligned} g_1(x)&=x+1+\frac2{x+1}&&\text{polynomial division}\\ \lim_{x\to\pm\infty}\frac{g_1(x)}x&=1 &&\text{candidate gradient}\\ \lim_{x\to\pm\infty}(g_1(x)-x)&=1 &&\text{candidate intercept}\\ g_1(x)-(x+1)&=\frac2{x+1}\longrightarrow0 &&\text{final verification} \end{aligned}

Why this step? The gradient limit alone does not determine the intercept. Subtract the full proposed line to verify it.

Answer. Oblique asymptote $y=x+1$ at both infinite ends.

Check. At $x=9$ , the value is $10.2$ , which is $0.2$ above the line; the residual decreases towards zero.

These equations also exclude $x=-1$ , where their non-zero remainders produce a vertical asymptote. A single curve can have both a vertical and an oblique asymptote.

Watch out

Wrong: call $y=x$ the asymptote merely because $g_1(x)/x\to1$ . Right: also find the offset; here $g_1(x)-x\to1$ , so the line is $y=x+1$ .

Change only the quadratic coefficient once more.

Approaching a curve does not prevent crossing it

The limiting comparison need not be a straight line. A graph can approach a parabola or another supplied curve while both grow without bound.

Asymptotic approach to another graph

f(x)-g(x)\longrightarrow0

along an infinite branch, the graph of $f$ approaches the graph $y=g(x)$ in vertical separation. For a non-linear $g$ , describe this as asymptotic approach to a curve rather than a straight-line asymptote.

In particular, for $|x|\to\infty$ ,

\left|\frac{\sin x}{x}\right|\le\frac1{|x|}\longrightarrow0

This permits oscillation and crossings while the separation shrinks.

Not in the formula booklet: learn it.

CAS can evaluate the difference limit directly. The bound explains why the output is zero despite the continuing oscillation.

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TC10–TC12 - Limits, asymptotes and cusps

A missing input need not give an infinite branch

Check the difference from a proposed line

Approaching a curve does not prevent crossing it

Continue this lesson.

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