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Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q1: 5x−35x-3 is continuous at 00⁠, −3-3 and 55

Exercise 5.1Continuity at a point · Section 5.2
The question

Prove that the function f(x)=5x−3f(x) = 5x - 3 is continuous at x=0,x = 0, at x=−3x = -3 and at x=5.x = 5.

Exam answer

Here f(x)=5x−3f(x) = 5x - 3⁠. Let x=cx = c be any real number.

LHL=lim⁡h→0f(c−h)=lim⁡h→0(5(c−h)−3)=5c−3RHL=lim⁡h→0f(c+h)=lim⁡h→0(5(c+h)−3)=5c−3\begin{aligned} \text{LHL} &= \lim_{h\to0} f(c - h) \\ &= \lim_{h\to0}\bigl(5(c - h) - 3\bigr) \\ &= 5c - 3 \\ \text{RHL} &= \lim_{h\to0} f(c + h) \\ &= \lim_{h\to0}\bigl(5(c + h) - 3\bigr) \\ &= 5c - 3 \end{aligned}

and f(c)=5c−3f(c) = 5c - 3⁠. So LHL=RHL=f(c)\text{LHL} = \text{RHL} = f(c)⁠, and ff is continuous at every real cc⁠.

At x=0x = 0⁠, −3-3⁠, 55 the common values are f(0)=−3f(0) = -3⁠, f(−3)=−18f(-3) = -18⁠, f(5)=22f(5) = 22⁠.

Hence ff is continuous at x=0x = 0⁠, at x=−3x = -3 and at x=5x = 5⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

A function is continuous at a point cc of its domain when three numbers agree: the left-hand limit (what f(x)f(x) approaches as xx comes towards cc from below), the right-hand limit (the same from above) and the value f(c).f(c). For 5x−3,5x - 3, moving xx by a tiny amount hh moves f(x)f(x) by only 5h,5h, which is tiny too. So all three numbers come out equal at every point, and in particular at 0,0, −3-3 and 5.5.

Given, to show, and the plan
Given
f(x)=5x−3,f(x) = 5x - 3, a formula that makes sense for every real x.x.
To show
ff is continuous at x=0,x = 0, at x=−3x = -3 and at x=5.x = 5.
Plan
(1) Check that each point is in the domain. (2) At each of the three points, compute the left-hand limit, the right-hand limit and the value separately, then compare. (3) Conclude, and note why the same works at every point.

Basics you need

1Continuity at a point

ff is continuous at a point cc of its domain if lim⁡x→cf(x)=f(c).\lim_{x\to c} f(x) = f(c). Spelled out: the left-hand limit, the right-hand limit and f(c)f(c) all exist and are equal (Definition 1, Section 5.2). Lesson: continuity at a point

2One-sided limits with a small step hh

Take h>0.h > 0. Points just to the left of cc are c−hc - h and points just to the right are c+h.c + h. So lim⁡x→c−f(x)=lim⁡h→0f(c−h),\lim_{x\to c^-} f(x) = \lim_{h\to 0} f(c-h), lim⁡x→c+f(x)=lim⁡h→0f(c+h).\lim_{x\to c^+} f(x) = \lim_{h\to 0} f(c+h). Basics: left-hand and right-hand limits

3A multiple of hh goes to 0

If h→0h \to 0 then 5h→05h \to 0 as well, because ∣5h∣=5∣h∣|5h| = 5|h| can be made as small as we like by taking hh small enough. So lim⁡h→0(a+5h)=a\lim_{h\to 0}(a + 5h) = a for any fixed number a.a. Basics: what a limit is

Solution

  1. The three points are in the domain

    5x−35x - 3 involves only multiplying by 5 and subtracting 3, which can be done for every real number. So the domain of ff is all of R,\R, and 0,0, −3,-3, 55 all belong to it.

    Continuity is only defined at points of the domain (Definition 1 starts with "let cc be a point in the domain of ff"). So before testing a point we confirm that ff is defined there.

  2. At x=0x = 0⁠: three numbers, computed separately

    Let h>0h > 0 be small. Then 0−h=−h0 - h = -h is just to the left of 0, and 0+h=h0 + h = h is just to the right.

    Left-hand limit.

    lim⁡x→0−f(x)=lim⁡h→0f(−h)=lim⁡h→0(5(−h)−3)=lim⁡h→0(−3−5h)=−3.\begin{aligned} \lim_{x\to 0^-} f(x) &= \lim_{h\to 0} f(-h)\\ &= \lim_{h\to 0} \bigl(5(-h) - 3\bigr)\\ &= \lim_{h\to 0} (-3 - 5h) = -3. \end{aligned}

    Right-hand limit.

    lim⁡x→0+f(x)=lim⁡h→0f(h)=lim⁡h→0(5h−3)=−3.\begin{aligned} \lim_{x\to 0^+} f(x) &= \lim_{h\to 0} f(h)\\ &= \lim_{h\to 0} (5h - 3) = -3. \end{aligned}

    Value. f(0)=5(0)−3=−3.f(0) = 5(0) - 3 = -3.

    All three equal −3,-3, so ff is continuous at x=0x = 0.

    The term −5h-5h (or 5h5h⁠) is the only part that depends on h,h, and it shrinks to 0 as hh does (Basic 3). What is left is −3,-3, on both sides. That common value is exactly f(0),f(0), which is what continuity asks for.
    LHL = RHL = value, all three. Basics: left-hand and right-hand limits

  3. At x=−3x = -3

    Now the nearby points are −3−h-3 - h (left) and −3+h-3 + h (right).

    Left-hand limit.

    lim⁡x→−3−f(x)=lim⁡h→0f(−3−h)=lim⁡h→0(5(−3−h)−3)=lim⁡h→0(−15−5h−3)=lim⁡h→0(−18−5h)=−18.\begin{aligned} \lim_{x\to -3^-} f(x) &= \lim_{h\to 0} f(-3-h)\\ &= \lim_{h\to 0} \bigl(5(-3-h) - 3\bigr)\\ &= \lim_{h\to 0} (-15 - 5h - 3)\\ &= \lim_{h\to 0} (-18 - 5h) = -18. \end{aligned}

    Right-hand limit.

    lim⁡x→−3+f(x)=lim⁡h→0f(−3+h)=lim⁡h→0(5(−3+h)−3)=lim⁡h→0(−18+5h)=−18.\begin{aligned} \lim_{x\to -3^+} f(x) &= \lim_{h\to 0} f(-3+h)\\ &= \lim_{h\to 0} \bigl(5(-3+h) - 3\bigr)\\ &= \lim_{h\to 0} (-18 + 5h) = -18. \end{aligned}

    Value. f(−3)=5(−3)−3=−15−3=−18.f(-3) = 5(-3) - 3 = -15 - 3 = -18.

    All three equal −18,-18, so ff is continuous at x=−3x = -3.

    The only place to slip is the sign: 5×(−3−h)=−15−5h,5 \times (-3 - h) = -15 - 5h, because the 5 multiplies both terms in the bracket. Keeping the bracket until the multiplication is done avoids the common error −15+5h-15 + 5h on the left side.

  4. At x=5x = 5

    Left-hand limit.

    lim⁡x→5−f(x)=lim⁡h→0f(5−h)=lim⁡h→0(25−5h−3)=lim⁡h→0(22−5h)=22.\begin{aligned} \lim_{x\to 5^-} f(x) &= \lim_{h\to 0} f(5-h)\\ &= \lim_{h\to 0} (25 - 5h - 3)\\ &= \lim_{h\to 0} (22 - 5h) = 22. \end{aligned}

    Right-hand limit.

    lim⁡x→5+f(x)=lim⁡h→0f(5+h)=lim⁡h→0(25+5h−3)=lim⁡h→0(22+5h)=22.\begin{aligned} \lim_{x\to 5^+} f(x) &= \lim_{h\to 0} f(5+h)\\ &= \lim_{h\to 0} (25 + 5h - 3)\\ &= \lim_{h\to 0} (22 + 5h) = 22. \end{aligned}

    Value. f(5)=25−3=22.f(5) = 25 - 3 = 22.

    All three equal 22, so ff is continuous at x=5x = 5.

    Same pattern again: at any point c,c, f(c±h)=(5c−3)±5h.f(c \pm h) = (5c - 3) \pm 5h. The ±5h\pm 5h dies away and leaves 5c−3=f(c).5c - 3 = f(c). That is why the three answers always match.

  5. Conclude, and see the general picture

    At each of x=0,x = 0, −3-3 and 55 the left-hand limit, the right-hand limit and the value of ff are equal, so ff is continuous at all three points, as required.

    In fact the calculation of Step 4 works at every real cc⁠: for h>0,h > 0,

    f(c−h)=5c−3−5h,f(c+h)=5c−3+5h,f(c - h) = 5c - 3 - 5h,\quad f(c + h) = 5c - 3 + 5h,

    and both tend to 5c−3=f(c).5c - 3 = f(c). So ff is continuous at every real number. This also follows at once from Example 14: every polynomial function is continuous, and 5x−35x - 3 is a polynomial.

    x y −4 −3 −2 −1 1 2 3 4 5 6 −20 −10 10 20 O (−3, −18) (5, 22) (0, −3) y = 5x − 3
    The graph of y=5x−3y = 5x - 3 is one unbroken straight line. The three points asked about, (0,−3),(0,-3), (−3,−18)(-3,-18) and (5,22),(5,22), sit on it with no gap or jump around them. (The two axes use different scales; read values from the tick labels.)
    The question names three points, so the proof must cover those three; checking them one by one is complete. Seeing that the same algebra works for a general cc shows the three points are not special: a straight line can be drawn without lifting the pen anywhere.
    Polynomials are continuous everywhere: Example 14. Lesson: continuity at a point
Result

f(x)=5x−3f(x) = 5x - 3 is continuous at x=0,x = 0, x=−3x = -3 and x=5x = 5⁠: at these points the left-hand limit, the right-hand limit and the value are −3,-3, −18-18 and 2222 respectively, each the same on both sides. (The answer key gives no answer for this "prove that" question.)

Check it with numbers

Take h=0.001.h = 0.001.

  • Near 0: f(−0.001)=−3.005f(-0.001) = -3.005 and f(0.001)=−2.995,f(0.001) = -2.995, both within 0.005 of f(0)=−3.f(0) = -3.
  • Near −3-3⁠: f(−3.001)=−18.005f(-3.001) = -18.005 and f(−2.999)=−17.995,f(-2.999) = -17.995, both within 0.005 of f(−3)=−18.f(-3) = -18.
  • Near 5: f(4.999)=21.995f(4.999) = 21.995 and f(5.001)=22.005,f(5.001) = 22.005, both within 0.005 of f(5)=22.f(5) = 22.

Each time the gap is 5h=0.005,5h = 0.005, exactly the ±5h\pm 5h term from the working, and it shrinks as hh does.

Common mistakes
  1. Checking only f(c).f(c). Being defined at a point is not the same as being continuous there. The limits from both sides must also equal f(c).f(c).
  2. Sign slips with c−h.c - h. 5(−3−h)=−15−5h,5(-3 - h) = -15 - 5h, not −15+5h.-15 + 5h. Multiply out the bracket before collecting terms.
  3. Writing f(c)=lim⁡f(x)f(c) = \lim f(x) as the starting point. That equality is what has to be shown; compute the two sides separately and only then compare them.
What to take away
  1. To test continuity at cc⁠: confirm cc is in the domain, then compute the left-hand limit, the right-hand limit and f(c)f(c) separately, and compare.
  2. Writing nearby points as c−hc - h and c+hc + h with h>0h > 0 turns one-sided limits into ordinary algebra in h.h.
  3. Polynomials (like 5x−35x - 3⁠) are continuous at every real number (Example 14).

Every fact used, and where it comes from

FactUsed inSource
Continuity at cc⁠: left-hand limit, right-hand limit and f(c)f(c) exist and are equalSteps 2–5Definition 1, Section 5.2 (Lesson: continuity at a point)
Continuity is tested only at points of the domainStep 1Definition 1, Section 5.2
Left and right limits as lim⁡h→0f(c∓h),\lim_{h\to0} f(c\mp h), h>0h > 0Steps 2–4Left and right hand limits, Class XI Limits (Basics: one-sided limits)
lim⁡h→0(a+kh)=a\lim_{h\to 0}(a + kh) = a for constants a,a, kkSteps 2–5Algebra of limits, Class XI Limits (Basics: what a limit is)
Every polynomial function is continuousStep 5Example 14, Section 5.2
Values in the check boxCheckRecomputed in the site's verification script
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