CEGEP Calculus I — Limits and Continuity Worksheet

All 11 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Limits from a Graph and a Table

   The graph of a function $f$ consists of three pieces: a line segment rising from $(-4,-1)$ to an open circle at $(-1,2)$; a solid dot at $(-1,3)$; and a smooth curve falling from an open circle at $(-1,2)$, through $(0,1)$, to $(3,-2)$.

   1. Find ${\displaystyle {lim}_{x\to -1^{-}}f(x)}$ and ${\displaystyle {lim}_{x\to -1^{+}}f(x)}$.
   2. Does ${\displaystyle {lim}_{x\to -1}f(x)}$ exist? Justify your answer from part (a).
   3. State $f(-1)$, and explain in one sentence why your answer to (b) is not affected by it.
2. Q2Limit Laws and Algebraic Evaluation

   Suppose ${\displaystyle {lim}_{x\to 2}f(x)=6}$ and ${\displaystyle {lim}_{x\to 2}g(x)=-3}$. Evaluate each limit, naming the limit law that justifies each step.

   1. ${\displaystyle {lim}_{x\to 2}[3f(x)+2g(x)]}$
   2. ${\displaystyle {lim}_{x\to 2}\frac{f(x)g(x)}{f(x)+g(x)}}$
   3. ${\displaystyle {lim}_{x\to 2}\sqrt{2f(x)+4}}$
3. Q3Indeterminate Forms and Algebraic Manipulation

   Evaluate each limit exactly, showing the manipulation that removes the indeterminate form.

   1. ${\displaystyle {lim}_{x\to 3}\frac{x^2-x-6}{x^2-9}}$
   2. ${\displaystyle {lim}_{x\to 0}\frac{\sqrt{x+4}-2}{x}}$
4. Q4One-Sided Limits and Piecewise Functions

   Two functions are given.

   1. For $f(x)={\displaystyle \frac{2x-10}{|x-5|}}$, $x\ne 5$, find ${\displaystyle {lim}_{x\to 5^{-}}f(x)}$ and ${\displaystyle {lim}_{x\to 5^{+}}f(x)}$, and decide whether ${\displaystyle {lim}_{x\to 5}f(x)}$ exists.
   2. For $$g(x)=\{\begin{array}{cc}{x}^2+1,\hfill & x<2,\hfill \\ [\mathrm{2pt}]7-x,\hfill & x\ge 2,\hfill \end{array}$$ find ${\displaystyle {lim}_{x\to 2^{-}}g(x)}$, ${\displaystyle {lim}_{x\to 2^{+}}g(x)}$, and ${\displaystyle {lim}_{x\to 2}g(x)}$ if it exists.
   3. State the condition on the one-sided limits under which a two-sided limit exists.
5. Q5Infinite Limits and Vertical Asymptotes

   Let $f(x)={\displaystyle \frac{x+1}{x^2-x-6}}$.

   1. Find the equations of the vertical asymptotes of $f$, and say what makes each one an asymptote rather than a hole.
   2. Evaluate the four one-sided limits at those two values of $x$.
6. Q6Limits at Infinity and Horizontal Asymptotes

   Evaluate each limit, then state the horizontal asymptotes of the function involved.

   1. ${\displaystyle {lim}_{x\to \infty }\frac{3x^2-5x}{2x^2+7}}$
   2. ${\displaystyle {lim}_{x\to \infty }\frac{4x+1}{x^3-2}}$
   3. ${\displaystyle {lim}_{x\to \infty }\frac{x^3+1}{2x^2-x}}$
7. Q7Limits at Infinity and Horizontal Asymptotes

   Let $f(x)={\displaystyle \frac{\sqrt{4x^2+3x}}{x+1}}$.

   1. Evaluate ${\displaystyle {lim}_{x\to \infty }f(x)}$.
   2. Evaluate ${\displaystyle {lim}_{x\to -\infty }f(x)}$.
   3. State every horizontal asymptote of $f$, and explain in one sentence why the two answers differ in sign.
8. Q8Continuity at a Point

   A function is defined by $$f(x)=\{\begin{array}{cc}{\displaystyle \frac{x^2-x-6}{x-3}},\hfill & x\ne 3,\hfill \\ [\mathrm{10pt}]4,\hfill & x=3.\hfill \end{array}$$

   1. State the three conditions required for $f$ to be continuous at $x=3$, and check each one.
   2. Classify the discontinuity.
   3. Change one number in the definition so that $f$ becomes continuous at $x=3$.
9. Q9Continuity on an Interval

   Let $f(x)={\displaystyle \frac{\sqrt{5-x}}{x^2-x-2}}$. Determine the largest intervals on which $f$ is continuous, naming the results about continuity that you use. Pay particular attention to the point $x=5$.

10. Q10The Intermediate Value Theorem

    Let $p(x)=x^3-4x+1$.

    1. Show that the equation $p(x)=0$ has a solution in the interval $(0,1)$, naming the theorem you use and checking each of its hypotheses.
    2. By evaluating $p$ twice more, locate a solution inside an interval of length $0.25$.
    3. Does your argument show that the solution is unique? Justify your answer.
11. Q11Synthesis — drawing on several topics in this set

    A holding tank at a desalination plant is monitored from $t=0$ hours. The salt concentration, in $\text{mg}/\text{L}$, is modelled by $$c(t)=\{\begin{array}{cc}{\displaystyle \frac{t^2-9}{t-3}},\hfill & 0\le t<3,\hfill \\ [\mathrm{10pt}]a,\hfill & t=3,\hfill \\ [\mathrm{6pt}]{\displaystyle \frac{18t+b}{t+3}},\hfill & t>3,\hfill \end{array}$$ where $a$ and $b$ are constants.

    1. Find the values of $a$ and $b$ that make $c$ continuous at $t=3$.
    2. With those values, evaluate ${\displaystyle {lim}_{t\to \infty }c(t)}$ and interpret the result in context, with units.
    3. Does the concentration ever actually reach the value found in (b)? Justify algebraically.

The 6 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

A limit is a claim about the neighbourhood, not about the point

Q1 gives you a graph in three pieces: a segment rising to an open circle, a solid dot sitting at a different height above the same input, and a curve leaving that open circle again. The marks are arranged that way deliberately, because three different quantities live at that one location and a graph answers them in three different places.

Both one-sided limits must exist and agree. Q4(c) asks you to state that condition outright, because it is the rule every later question in the set appeals to: parts (a) and (b) of the same question are two situations where it is tested and only one of them passes.

Naming the law is the answer, not decoration

Q2 hands you two limits and asks for three combinations, naming the limit law that justifies each step. The arithmetic is a few seconds' work; the marks are in the naming, because two of the three have a hypothesis that has to be verified before the law may be quoted at all.

Every one of these laws runs in one direction only: if the pieces have limits, then the combination does. Nothing in them licenses the converse, so a combination with a perfectly good limit is no evidence at all that its pieces have any.

0/0 is a question. Anything else is already an answer

Substitute first, always — it costs one line and it tells you which case you are in. Q3 is two limits that both return 0/0, and the form is indeterminate precisely because it does not determine the answer: different functions producing it have different limits. What the expression looks like tells you which rewrite is needed.

An absolute value or a piecewise rule means two calculations

Q4 puts the two commonest sources of a side-dependent limit side by side. In part (a) the absolute value in the denominator has to be unwrapped by side before anything can be simplified: on one side the bars come off unchanged, on the other they come off with a sign, and the two resulting expressions are genuinely different functions.

A zero of the denominator is a candidate, not a verdict

Q5 asks for the vertical asymptotes of a rational function and for what makes each one an asymptote rather than a hole. Those are two different findings and the second is where the reasoning is.

Part (b) then asks for four one-sided limits, and the technique is sign analysis rather than substitution. Establish that the numerator tends to a non-zero number while the denominator tends to 0, then decide the sign of the denominator just to the left and just to the right — with a factored denominator that means asking what sign each factor carries near the point. A small negative denominator under a positive numerator gives −∞; a small positive one gives +∞. The two sides routinely disagree, which is why the question asks for all four separately.

Two ends of a graph are two questions

This pair of sentences is worth memorising, because the two objects are constantly conflated: a vertical asymptote is an infinite limit at a finite input; a horizontal asymptote is a finite limit at infinity. Different limit, different question, different method.

For Q6 the method is uniform: divide numerator and denominator by the highest power of x in the denominator, then let every term with an x underneath go to 0. The three parts are the three degree cases on purpose — one where the degrees match, one where the numerator is outgunned, and one where the numerator wins and there is no horizontal asymptote at all. Reporting an asymptote in that last case is the trap; the honest answer names the behaviour instead.

Continuity at a point: three conditions, and the failure classifies itself

Q8 asks you to state the three conditions and check each one. Learn them as a numbered list, because naming a discontinuity is nothing more than recording which condition broke.

Continuity on an interval, and the endpoint that decides the bracket

Q9 asks for the largest intervals on which a function is continuous, naming the results you use. The theorems are short — a polynomial is continuous everywhere; a root is continuous at every point of its domain; a quotient of continuous functions is continuous wherever the denominator is non-zero — so all the work is in the domain. Find where the radicand is non-negative, remove the zeros of the denominator, and report what is left.

What the Intermediate Value Theorem actually promises

Q10 asks you to name the theorem and check each of its hypotheses. There are exactly two, and both have to be written down.

Part (b) turns the theorem into a method. Evaluate at the midpoint, see which half now has endpoint values straddling the target, and apply the theorem again on that half; each pass halves the interval, so reaching a stated width is just a matter of counting the passes. Part (c) then asks whether you have shown the solution is unique, and the answer is the part students most often get backwards.

The synthesis question, and why it is last

Q11 is a modelled quantity defined by three rules — one indeterminate quotient before a changeover time, a single assigned value at it, and a rational rule after — with two unknown constants in it. Nothing in it is new. Part (a) is the three conditions for continuity used as a pair of equations: take the one-sided limit on the left (which needs the factoring from Q3), take the one-sided limit on the right, and force both to agree with the assigned value. Part (b) is a limit at infinity read back into the situation, with units, as a long-run levelling off. Part (c) asks whether that level is ever actually reached, and the honest way to answer is algebraic: rewrite the rule so the gap between the function and its asymptote is a single visible term, and show that term is never zero. An asymptote is a value approached, not a value attained, and this is the question that makes the difference concrete.

Getting the most out of it

Substitute before you do anything clever

Most limits on this sheet are settled or classified by putting the number in. One line tells you which of three cases you are in: a real number means you are finished; 0/0 means rewrite; a non-zero number over 0 means the quotient is unbounded and the real question is the sign on each side. Only one of the three needs algebra.

Write the restriction every time you cancel

"= x + 2 for x ≠ 3" is not pedantry — it is the sentence that makes the cancellation legal, and it is what keeps the hole in your picture of the graph. The students who drop the clause are the ones who later call a function continuous at a point that is not in its domain.

Name the law, the condition or the theorem out loud

Several questions here award the mark for the justification rather than the value: the limit law at the decisive step and its hypothesis, the three conditions of continuity one at a time, the continuity theorems behind an interval answer, the two hypotheses of the IVT. Practise writing the reason as a clause — "by the quotient law, and the denominator's limit is non-zero", "continuous on [0, 1] because it is a polynomial". The devis asks for rigorous mathematical reasoning, and on a CEGEP final that clause is where it is looked for.

Take the sides seriously

Absolute values, piecewise rules, vertical asymptotes and square roots at the two ends of a graph are four different topics with one habit in common: the answer depends on which side of the point, or which end of the axis, you are on. Whenever one appears, do the calculation twice and compare — the comparison is usually the thing being marked.

Common questions