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Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q2: Continuity of 2x2−12x^2-1 at x=3x=3

Exercise 5.1Continuity at a point · Section 5.2
The question

Examine the continuity of the function f(x)=2x2−1f(x) = 2x^2 - 1 at x=3.x = 3.

Exam answer

Here f(x)=2x2−1f(x) = 2x^2 - 1⁠. At x=3x = 3⁠:

LHL=lim⁡h→0f(3−h)=lim⁡h→0(2(3−h)2−1)=17RHL=lim⁡h→0f(3+h)=lim⁡h→0(2(3+h)2−1)=17\begin{aligned} \text{LHL} &= \lim_{h\to0} f(3 - h) \\ &= \lim_{h\to0}\bigl(2(3 - h)^2 - 1\bigr) \\ &= 17 \\ \text{RHL} &= \lim_{h\to0} f(3 + h) \\ &= \lim_{h\to0}\bigl(2(3 + h)^2 - 1\bigr) \\ &= 17 \end{aligned}

and f(3)=2(3)2−1=17f(3) = 2(3)^2 - 1 = 17⁠. So LHL=RHL=f(3)\text{LHL} = \text{RHL} = f(3)⁠.

Hence ff is continuous at x=3x = 3⁠.

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

The one idea this problem is built on

"Examine the continuity at x=3x = 3" means: find the left-hand limit, the right-hand limit and the value f(3),f(3), and see whether all three agree. Expanding f(3±h)f(3 \pm h) shows that ff changes by only ±12h+2h2\pm 12h + 2h^2 when xx moves by h,h, and that change vanishes as h→0.h \to 0. So all three numbers equal f(3)=17f(3) = 17 and ff is continuous at 3.

Given, to find, and the plan
Given
f(x)=2x2−1,f(x) = 2x^2 - 1, defined for every real x.x.
To find
Whether ff is continuous at x=3.x = 3.
Plan
(1) Check 3 is in the domain and find f(3).f(3). (2) Left-hand limit through f(3−h).f(3 - h). (3) Right-hand limit through f(3+h).f(3 + h). (4) Compare.

Basics you need

1Continuity at a point

ff is continuous at cc (a point of its domain) if 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

With h>0h > 0⁠: the left-hand limit at cc is lim⁡h→0f(c−h)\lim_{h\to0} f(c-h) and the right-hand limit is lim⁡h→0f(c+h).\lim_{h\to0} f(c+h). Basics: left-hand and right-hand limits

3Squaring a sum or difference

(a±b)2=a2±2ab+b2.(a \pm b)^2 = a^2 \pm 2ab + b^2. So (3±h)2=9±6h+h2.(3 \pm h)^2 = 9 \pm 6h + h^2.

4Terms in hh and h2h^2 go to 0

As h→0,h \to 0, both 12h→012h \to 0 and 2h2→02h^2 \to 0 (for ∣h∣<1,|h| < 1, h2h^2 is even smaller than ∣h∣|h|⁠). Basics: what a limit is

Solution

  1. 3 is in the domain; find f(3)f(3)

    2x2−12x^2 - 1 can be computed for every real x,x, so the domain is R\R and 33 belongs to it.

    f(3)=2(3)2−1=2×9−1=17.f(3) = 2(3)^2 - 1 = 2 \times 9 - 1 = 17.
    Definition 1 only speaks about points in the domain, so this is the first thing to confirm. The value f(3)f(3) is the target the two limits must hit.

  2. Left-hand limit at 3

    Let h>0.h > 0. The point 3−h3 - h lies just to the left of 3.

    f(3−h)=2(3−h)2−1=2(9−6h+h2)−1=18−12h+2h2−1=17−12h+2h2.\begin{aligned} f(3-h) &= 2(3-h)^2 - 1\\ &= 2(9 - 6h + h^2) - 1\\ &= 18 - 12h + 2h^2 - 1\\ &= 17 - 12h + 2h^2. \end{aligned}

    As h→0,h \to 0, −12h→0-12h \to 0 and 2h2→0,2h^2 \to 0, so

    lim⁡x→3−f(x)=lim⁡h→0(17−12h+2h2)=17.\lim_{x\to 3^-} f(x) = \lim_{h\to 0}(17 - 12h + 2h^2) = 17.
    Expanding the square separates f(3−h)f(3-h) into a fixed part (17) and parts that carry a factor of h.h. Only the fixed part survives the limit. This is the whole mechanism of continuity for a polynomial.
    (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2 with a=3,a = 3, b=h.b = h. Basics: left-hand and right-hand limits

  3. Right-hand limit at 3

    The point 3+h3 + h (with h>0h > 0⁠) lies just to the right of 3.

    f(3+h)=2(3+h)2−1=2(9+6h+h2)−1=17+12h+2h2.\begin{aligned} f(3+h) &= 2(3+h)^2 - 1\\ &= 2(9 + 6h + h^2) - 1\\ &= 17 + 12h + 2h^2. \end{aligned} lim⁡x→3+f(x)=lim⁡h→0(17+12h+2h2)=17.\lim_{x\to 3^+} f(x) = \lim_{h\to 0}(17 + 12h + 2h^2) = 17.
    The two sides differ only in the sign of the 12h12h term, and that term disappears in the limit either way. So approaching from below and from above leads to the same number.

  4. Compare the three numbers

    lim⁡x→3−f(x)=17,lim⁡x→3+f(x)=17,f(3)=17.\lim_{x\to 3^-} f(x) = 17,\quad \lim_{x\to 3^+} f(x) = 17,\quad f(3) = 17.

    All three are equal, so ff is continuous at x=3x = 3.

    This is exactly Definition 1. As a cross-check, 2x2−12x^2 - 1 is a polynomial, and Example 14 shows every polynomial is continuous at every real number, so the answer had to be "continuous".
    LHL = RHL = f(c)f(c) means continuous. Lesson: continuity at a point
Result

ff is continuous at x=3.x = 3. The left-hand limit, the right-hand limit and f(3)f(3) all equal 17. (Textbook answer: "ff is continuous at x=3x = 3".)

Check it with numbers

With h=0.001h = 0.001⁠: f(2.999)=2(8.994001)−1f(2.999) = 2(8.994001) - 1 =16.988002= 16.988002 and f(3.001)=2(9.006001)−1f(3.001) = 2(9.006001) - 1 =17.012002.= 17.012002. The formulas give 17−0.012+0.000002=16.98800217 - 0.012 + 0.000002 = 16.988002 ✓ and 17+0.012+0.000002=17.01200217 + 0.012 + 0.000002 = 17.012002 ✓. Both are within about 0.012 of f(3)=17,f(3) = 17, and the gap shrinks with h.h.

Common mistakes
  1. (3−h)2=9−h2.(3 - h)^2 = 9 - h^2. The middle term −6h-6h is missing; the square of a difference has three terms.
  2. Forgetting to multiply every term by 2. 2(9−6h+h2)=18−12h+2h2.2(9 - 6h + h^2) = 18 - 12h + 2h^2.
  3. Stopping at "f(3)=17f(3) = 17 exists". A value at the point is only one of the three numbers; both one-sided limits are needed too.
What to take away
  1. Continuity at cc⁠: LHL, RHL and f(c),f(c), computed separately, must agree.
  2. For a polynomial, f(c±h)f(c \pm h) is f(c)f(c) plus terms with a factor of h,h, which vanish as h→0.h \to 0. That is why polynomials are continuous everywhere.

Every fact used, and where it comes from

FactUsed inSource
Continuity at cc⁠: LHL, RHL and f(c)f(c) exist and are equalSteps 1, 4Definition 1, Section 5.2 (Lesson: continuity at a point)
Left and right limits as lim⁡h→0f(c∓h),\lim_{h\to0} f(c\mp h), h>0h > 0Steps 2, 3Class XI Limits (Basics: one-sided limits)
(a±b)2=a2±2ab+b2(a \pm b)^2 = a^2 \pm 2ab + b^2Steps 2, 3Algebraic identity
lim⁡h→0(a+bh+ch2)=a\lim_{h\to0}(a + bh + ch^2) = aSteps 2, 3Algebra of limits, Class XI Limits (Basics: what a limit is)
Every polynomial function is continuousStep 4Example 14, Section 5.2
Values in the check boxCheckRecomputed in the site's verification script
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