Navigation

Limits

2.1.1 Calculating limits with limit laws

48 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Suppose lim⁡x→af(x)=0\displaystyle\lim_{x \rightarrow a} f(x)=0 and lim⁡x→ag(x)=0\displaystyle\lim_{x \rightarrow a} g(x)=0. Which of the following limits can you compute, given this information?

  1. lim⁡x→af(x)2\displaystyle\lim_{x \rightarrow a} \frac{f(x)}{2}

  2. lim⁡x→a2f(x)\displaystyle\lim_{x \rightarrow a} \frac{2}{f(x)}

  3. lim⁡x→af(x)g(x)\displaystyle\lim_{x \rightarrow a} \frac{f(x)}{g(x)}

  4. lim⁡x→af(x)g(x)\displaystyle\lim_{x \rightarrow a} f(x)g(x)

Answer

(a) and (d)

Full solution
Q2Stage 1

Give two functions f(x)f(x) and g(x)g(x) that satisfy lim⁡x→3f(x)=lim⁡x→3g(x)=0\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0 and lim⁡x→3f(x)g(x)=10\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=10.

Hint

Try to make two functions with factors that will cancel.

Answer

There are many possible answers; one is f(x)=10(x−3)f(x)=10(x-3), g(x)=x−3g(x)=x-3.

Full solution
Q3Stage 1

Give two functions f(x)f(x) and g(x)g(x) that satisfy lim⁡x→3f(x)=lim⁡x→3g(x)=0\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0 and lim⁡x→3f(x)g(x)=0\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=0.

Hint

Try to make g(x)g(x) cancel out.

Answer

There are many possible answers; one is f(x)=(x−3)2f(x)=(x-3)^2 and g(x)=x−3g(x)=x-3. Another is f(x)=0f(x)=0 and g(x)=x−3g(x)=x-3.

Full solution
Q4Stage 1

Give two functions f(x)f(x) and g(x)g(x) that satisfy lim⁡x→3f(x)=lim⁡x→3g(x)=0\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0 and lim⁡x→3f(x)g(x)=∞\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=\infty.

Answer

There are many possible answers; one is f(x)=x−3f(x)=x-3, g(x)=(x−3)3g(x)=(x-3)^3.

Full solution
Q5Stage 1

Suppose lim⁡x→af(x)=lim⁡x→ag(x)=0\displaystyle\lim_{x \rightarrow a}f(x)=\displaystyle\lim_{x \rightarrow a}g(x)=0. What are the possible values of lim⁡x→af(x)g(x)\displaystyle\lim_{x \rightarrow a}\dfrac{f(x)}{g(x)}?

Hint

See Questions 2, 3, and 4.

Answer

Any real number; positive infinity; negative infinity; does not exist.

Full solution

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

For Questions 6 through 38, evaluate the given limits.

Q6Stage 2

lim⁡t→102(t−10)2t\displaystyle\lim_{t \rightarrow 10} \dfrac{2(t-10)^2}{t}

Hint

Find the limit of the numerator and denominator separately.

Answer

0

Full solution
Q7Stage 2

lim⁡y→0(y+1)(y+2)(y+3)cos⁡y\displaystyle\lim_{y \rightarrow 0} \dfrac{(y+1)(y+2)(y+3)}{\cos y}

Hint

Break it up into smaller pieces, evaluate the limits of the pieces.

Answer

6

Full solution
Q8Stage 2

lim⁡x→3(4x−2x+2)4\displaystyle\lim_{x \rightarrow 3} \left(\dfrac{4x-2}{x+2}\right)^4

Hint

First find the limit of the “inside" function, 4x−2x+2\dfrac{4x-2}{x+2}.

Answer

16

Full solution
Q9Stage 2Past exam · 2015Q

lim⁡t→−3(1−tcos⁡(t))\ds \lim_{t\to -3} \left(\frac{1-t}{\cos(t)}\right)

Hint

Is cos⁡(−3)\cos(-3) zero?

Answer

4/cos⁡(3)4/\cos(3)

Full solution
Q10Stage 2Past exam · 2015Q

lim⁡h→0(2+h)2−42h\ds \lim_{h \to 0} \frac{(2+h)^2-4}{2h}

Hint

Expand, then simplify.

Answer

22

Full solution
Q11Stage 2Past exam · 2015Q

lim⁡t→−2(t−5t+4)\ds \lim_{t\to -2} \left(\frac{t-5}{t+4}\right)

Answer

−7/2-7/2

Full solution
Q12Stage 2Past exam · 2015Q

lim⁡x→15x3+4\ds \lim_{x\to 1} \sqrt{5x^3 + 4}

Answer

3

Full solution
Q13Stage 2Past exam · 2015Q

lim⁡t→−1(t−2t+3)\displaystyle\lim_{t\rightarrow -1} \left(\frac{t-2}{t+3}\right)

Answer

−32-\frac{3}{2}

Full solution
Q14Stage 2Past exam · 2012H

lim⁡x→1log⁡(1+x)−xx2\lim\limits_{x\rightarrow 1}\dfrac{\log(1+x)-x}{x^2}

Hint

Try the simplest method first.

Answer

log⁡(2)−1\log(2)-1

Full solution
Q15Stage 2Past exam · 2015Q

lim⁡x→2(x−2x2−4)\displaystyle\lim_{x\rightarrow 2} \left(\frac{x-2}{x^2-4}\right)

Hint

Factor the denominator.

Answer

14\frac{1}{4}

Full solution
Q16Stage 2Past exam · 2006H

lim⁡x→4x2−4xx2−16\ds\lim\limits_{x\rightarrow 4}\dfrac{x^2-4x}{x^2-16}

Hint

Factor the numerator and the denominator.

Answer

12\dfrac{1}{2}

Full solution
Q17Stage 2Past exam · 2007H

lim⁡x→2x2+x−6x−2\lim\limits_{x\rightarrow 2}\dfrac{x^2+x-6}{x-2}

Hint

Factor the numerator.

Answer

55

Full solution
Q18Stage 2Past exam · 2015Q

lim⁡x→−3x2−9x+3\ds \lim_{x \to -3} \frac{x^2-9}{x+3}

Hint

Simplify first by factoring the numerator.

Answer

−6-6

Full solution
Q19Stage 2

lim⁡t→212t4−3t3+t\displaystyle\lim_{t \rightarrow 2} \frac{1}{2}t^4-3t^3+t

Hint

The function is a polynomial.

Answer

−14-14

Full solution
Q20Stage 2Past exam · 2015Q

lim⁡x→−1x2+8−3x+1\ds \lim_{x\to -1} \frac{\sqrt{x^2+8}-3}{x+1}.

Hint

Multiply both the numerator and the denominator by the conjugate of the numerator, x2+8+3\sqrt{x^2+8}+3.

Answer

−13-\frac{1}{3}

Full solution
Q21Stage 2Past exam · 2015Q

lim⁡x→1x+2−4−xx−1\displaystyle \lim_{x\rightarrow 1} \frac{\sqrt{x+2}-\sqrt{4-x}}{x-1}

Hint

Multiply both the numerator and the denominator by the conjugate of the numerator, x+2+4−x\sqrt{x+2}+\sqrt{4-x}.

Answer

13\frac{1}{\sqrt{3}}

Full solution
Q22Stage 2Past exam · 2015Q

lim⁡x→3x−2−4−xx−3\ds \lim_{x\to 3} \frac{\sqrt{x-2}-\sqrt{4-x}}{x-3}.

Hint

Multiply both the numerator and the denominator by the conjugate of the numerator, x−2+4−x\sqrt{x-2}+\sqrt{4-x}.

Answer

1

Full solution
Q23Stage 2

lim⁡x→0−x2cos⁡(3x)\displaystyle\lim_{x \rightarrow 0}-x^2\cos\left(\frac{3}{x}\right)

Hint

Consider the factors x2x^2 and cos⁡(3x)\cos\left(\frac{3}{x}\right) separately.

Answer

0

Full solution
Q24Stage 2Past exam · 2012H

lim⁡x→0xsin⁡2(1x)\lim\limits_{x\rightarrow 0}x\sin^2\left(\dfrac{1}{x}\right)

Hint

Compare to the previous question.

Answer

0

Full solution
Q25Stage 2

lim⁡w→52w2−50(w−5)(w−1)\displaystyle\lim_{w \rightarrow 5} \dfrac{2w^2-50}{(w-5)(w-1)}

Hint

Factor the numerator.

Answer

5

Full solution
Q26Stage 2

lim⁡r→−5rr2+10r+25\displaystyle\lim_{r \rightarrow -5} \dfrac{r}{r^2+10r+25}

Hint

Factor the denominator; pay attention to signs.

Answer

−∞-\infty

Full solution
Q27Stage 2

lim⁡x→−1x3+x2+x+13x+3\displaystyle\lim_{x \rightarrow -1}\sqrt{\dfrac{x^3+x^2+x+1}{3x+3}}

Hint

First find the limit of the “inside" function.

Answer

23\sqrt{\dfrac{2}{3}}

Full solution
Q28Stage 2

lim⁡x→0x2+2x+13x5−5x3\displaystyle\lim_{x \rightarrow 0} \dfrac{x^2+2x+1}{3x^5-5x^3}

Hint

Factor; pay attention to signs.

Answer

DNE

Full solution
Q29Stage 2

lim⁡t→7t2x2+2tx+1t2−14t+49\displaystyle\lim_{t \rightarrow 7} \dfrac{t^2x^2+2tx+1}{t^2-14t+49}, where xx is a positive constant

Hint

Look for perfect squares

Answer

∞\infty

Full solution
Q30Stage 2

lim⁡d→0x5−32x+15\displaystyle\lim_{d \rightarrow 0} x^5-32x+15, where xx is a constant

Hint

Think about what effect changing dd has on the function x5−32x+15x^5-32x+15.

Answer

x5−32x+15x^5-32x+15

Full solution
Q31Stage 2

lim⁡x→1(x−1)2sin⁡[(x2−3x+2x2−2x+1)2+15]\displaystyle\lim_{x \rightarrow 1} (x-1)^2\sin\left[\left(\dfrac{x^2-3x+2}{x^2-2x+1}\right)^2+15\right]

Hint

There's an easy way.

Answer

00

Full solution
Q32Stage 2Past exam · 2006H

Evaluate

lim⁡x→0x1/101sin⁡(x−100)\lim_{x\rightarrow 0} x^{1/101} \sin\big(x^{-100}\big)

or explain why this limit does not exist.

Hint

What can you do to safely ignore the sine function?

Answer

00

Full solution
Q33Stage 2Past exam · 1999H

lim⁡x→2x2−4x2−2x\lim\limits_{x\rightarrow2}\dfrac{x^2-4}{x^2-2x}

Hint

Factor

Answer

2

Full solution
Q34Stage 2

lim⁡x→5(x−5)2x+5\displaystyle\lim_{x \rightarrow 5} \dfrac{(x-5)^2}{x+5}

Hint

If you're looking at the hints for this one, it's probably easier than you think.

Answer

0

Full solution
Q35Stage 2

Evaluate lim⁡t→1213t2+1t2−12t−1\ds\lim_{t \to \frac{1}{2}}\dfrac{\frac{1}{3t^2}+\frac{1}{t^2-1}}{2t-1} .

Hint

You'll want to simplify this, since t=12t=\frac{1}{2} is not in the domain of the function. One way to start your simplification is to add the fractions in the numerator by finding a common denominator.

Answer

−329-\dfrac{32}{9}

Full solution
Q36Stage 2

Evaluate lim⁡x→0(3+∣x∣x)\ds\lim_{x \to 0}\left( 3+\dfrac{|x|}{x}\right).

Hint

If you're not sure how ∣x∣x\dfrac{|x|}{x} behaves, try plugging in a few values of xx, like x=±1x=\pm 1 and x=±2x=\pm 2.

Answer

DNE

Full solution
Q37Stage 2

Evaluate lim⁡d→−4∣3d+12∣d+4\ds\lim_{d \to -4}\dfrac{|3d+12|}{d+4}

Hint

Look to Question 36 to see how a function of the form ∣X∣X\dfrac{|X|}{X} behaves.

Answer

DNE

Full solution
Q38Stage 2

Evaluate lim⁡x→05x−9∣x∣+2\ds\lim_{x \to 0}\dfrac{5x-9}{|x|+2}.

Hint

Is anything weird happening to this function at x=0x=0?

Answer

−92-\dfrac{9}{2}

Full solution
Q39Stage 2

Suppose lim⁡x→−1f(x)=−1\displaystyle\lim_{x \rightarrow -1} f(x)=-1. Evaluate lim⁡x→−1xf(x)+32f(x)+1\displaystyle\lim_{x \rightarrow -1} \dfrac{xf(x)+3}{2f(x)+1}.

Hint

Use the limit laws.

Answer

−4-4

Full solution
Q40Stage 2Past exam · 2007H

Find the value of the constant aa for which lim⁡x→−2x2+ax+3x2+x−2\lim\limits_{x\rightarrow -2}\dfrac{x^2+ax+3}{x^2+x-2} exists.

Hint

The denominator goes to zero; what must the numerator go to?

Answer

a=72a=\dfrac{7}{2}

Full solution
Q41Stage 2

Suppose f(x)=2xf(x)=2x and g(x)=1xg(x)=\frac{1}{x}. Evaluate the following limits.

  1. lim⁡x→0f(x)\displaystyle\lim_{x \rightarrow 0} f(x)

  2. lim⁡x→0g(x)\displaystyle\lim_{x \rightarrow 0} g(x)

  3. lim⁡x→0f(x)g(x)\displaystyle\lim_{x \rightarrow 0} f(x)g(x)

  4. lim⁡x→0f(x)g(x)\displaystyle\lim_{x \rightarrow 0} \dfrac{f(x)}{g(x)}

  5. lim⁡x→2[f(x)+g(x)]\displaystyle\lim_{x \rightarrow 2} [f(x)+g(x)]

  6. lim⁡x→0f(x)+1g(x+1)\displaystyle\lim_{x \rightarrow 0} \dfrac{f(x)+1}{g(x+1)}

Answer
  1. lim⁡x→0f(x)=0\displaystyle\lim_{x \rightarrow 0} f(x)=0

  2. lim⁡x→0g(x)=\displaystyle\lim_{x \rightarrow 0} g(x)= DNE

  3. lim⁡x→0f(x)g(x)=2\displaystyle\lim_{x \rightarrow 0} f(x)g(x)=2

  4. lim⁡x→0f(x)g(x)=0\displaystyle\lim_{x \rightarrow 0} \dfrac{f(x)}{g(x)}=0

  5. lim⁡x→2f(x)+g(x)=92\displaystyle\lim_{x \rightarrow 2} f(x)+g(x)=\dfrac{9}{2}

  6. lim⁡x→0f(x)+1g(x+1)=1\displaystyle\lim_{x \rightarrow 0} \dfrac{f(x)+1}{g(x+1)}=1

Full solution

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q42Stage 3Past exam · 2015Q

lim⁡x→2x+7−11−x2x−4\ds \lim_{x\to 2} \frac{\sqrt{x+7}-\sqrt{11-x}}{2x-4}.

Hint

Multiply both the numerator and the denominator by the conjugate of the numerator, x+7+11−x\sqrt{x+7}+\sqrt{11-x}.

Answer

16\frac{1}{6}

Full solution
Q43Stage 3Past exam · 2015Q

lim⁡t→13t−32−5−t\ds \lim_{t\to 1} \frac{3t-3}{2 - \sqrt{5-t}}.

Hint

Multiply both the numerator and the denominator by the conjugate of the denominator, 2+5−t2+\sqrt{5-t}.

Answer

12

Full solution
Q44Stage 3

The curve y=f(x)y=f(x) is shown in the graph below. Sketch the graph of y=1f(x)y=\dfrac{1}{f(x)}.

Figure from prob_s1.4, line 2

Figure from prob_s1.4, line 2

Hint

Try plotting points. If you can't divide by f(x)f(x), take a limit.

Answer

Figure from prob_s1.4, line 2

Figure from prob_s1.4, line 2

Pictures may vary somewhat; the important points are the values of the function at integer values of xx, and the vertical asymptotes.

Full solution
Q45Stage 3

The graphs of functions f(x)f(x) and g(x)g(x) are shown in the graphs below. Use these to sketch the graph of f(x)g(x)\dfrac{f(x)}{g(x)}.

Figure from prob_s1.4, line 2

Figure from prob_s1.4, line 2

Figure from prob_s1.4, line 10

Figure from prob_s1.4, line 10

Hint

There is a close relationship between ff and gg. Fill in the following table:

xxf(x)f(x)g(x)g(x)f(x)g(x)\dfrac{f(x)}{g(x)}
−3-3
−2-2
−1-1
−0-0
11
22
33
Answer

Figure from prob_s1.4, line 2

Figure from prob_s1.4, line 2

Full solution
Q46Stage 3

Let f(x)=1xf(x) = \frac{1}{x} and g(x)=−1xg(x) = \frac{-1}{x}.

  1. Evaluate lim⁡x→0f(x)\displaystyle\lim_{x \rightarrow 0} f(x) and lim⁡x→0g(x)\displaystyle\lim_{x \rightarrow 0} g(x).

  2. Evaluate lim⁡x→0[f(x)+g(x)]\displaystyle\lim_{x \rightarrow 0} [f(x)+g(x)]

  3. Is it always true that lim⁡x→a[f(x)+g(x)]=lim⁡x→af(x)+lim⁡x→ag(x)\displaystyle\lim_{x \rightarrow a} [f(x)+g(x)]= \displaystyle\lim_{x \rightarrow a} f(x)+\displaystyle\lim_{x \rightarrow a} g(x)?

Answer

(a) DNE , DNE (b) 0 (c) No: it is only true when both lim⁡x→af(x)\displaystyle\lim_{x \rightarrow a} f(x) and lim⁡x→ag(x)\displaystyle\lim_{x \rightarrow a} g(x) exist.

Full solution
Q47Stage 3

Suppose

f(x)={x2+3,x>00,x=0x2−3,x<0f(x)=\left\{\begin{array}{lcc} x^2+3&,&x>0\\ 0&,&x=0\\ x^2-3&,&x<0 \end{array}\right.
  1. Evaluate lim⁡x→0−f(x)\ds\lim_{x \to 0^-} f(x).

  2. Evaluate lim⁡x→0+f(x)\ds\lim_{x \to 0^+} f(x).

  3. Evaluate lim⁡x→0f(x)\ds\lim_{x \to 0} f(x).

Hint

When you're evaluating lim⁡x→0−f(x)\ds\lim_{x \to 0^-}f(x), you're only considering values of xx that are less than 0.

Answer

(a) lim⁡x→0−f(x)=−3\ds\lim_{x \to 0^-} f(x)=-3 (b) lim⁡x→0+f(x)=3\ds\lim_{x \to 0^+} f(x)=3 (c) lim⁡x→0f(x)=\ds\lim_{x \to 0} f(x)= DNE

Full solution
Q48Stage 3

Suppose

f(x)={x2+8x+16x2+30x−4,x>−4x3+8x2+16x,x≤−4f(x)=\left\{\begin{array}{lcc} \dfrac{x^2+8x+16}{x^2+30x-4}&,&x>-4\\ &\\ x^3+8x^2+16x&,&x\le-4 \end{array}\right.
  1. Evaluate lim⁡x→−4−f(x)\ds\lim_{x \to -4^-} f(x).

  2. Evaluate lim⁡x→−4+f(x)\ds\lim_{x \to -4^+} f(x).

  3. Evaluate lim⁡x→−4f(x)\ds\lim_{x \to -4} f(x).

Hint

When you're considering lim⁡x→−4−f(x)\ds\lim_{x \to -4^-}f(x), you're only considering values of xx that are less than −4-4.

When you're considering lim⁡x→−4+f(x)\ds\lim_{x \to -4^+}f(x), think about the domain of the rational function in the top line.

Answer

(a) lim⁡x→−4−f(x)=0\ds\lim_{x \to -4^-} f(x)=0 (b) lim⁡x→−4+f(x)=0\ds\lim_{x \to -4^+} f(x)=0 (c) lim⁡x→−4f(x)=0\ds\lim_{x \to -4} f(x)=0

Full solution

My list

nothing marked yet

Loading…

Open the whole list →

Your tutor can open this list with you. It follows your account, so it is there on whichever device you study on.

From the UBC Math 100 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.