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Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q20: Is x2−sin⁡x+5x^2-\sin x+5 continuous at x=πx=\pi⁠?

Exercise 5.1Algebra of continuous functions · Section 5.2.1
The question

20. Is the function defined by f(x)=x2−sin⁡x+5f(x)=x^2-\sin x+5 continuous at x=πx=\pi⁠?

Exam answer

f(π)=π2−sin⁡π+5=π2+5f(\pi) = \pi^2 - \sin\pi + 5 = \pi^2 + 5⁠.

Put x=π+hx = \pi + h⁠; as x→πx \to \pi⁠, h→0h \to 0⁠. Also sin⁡(π+h)=−sin⁡h\sin(\pi + h) = -\sin h⁠.

lim⁡x→πf(x)=lim⁡h→0f(π+h)=lim⁡h→0[(π+h)2+sin⁡h+5]=π2+5=f(π)\begin{aligned} \lim_{x\to\pi}f(x) &= \lim_{h\to0}f(\pi + h) \\ &= \lim_{h\to0}\big[(\pi + h)^2 \\ &\qquad + \sin h + 5\big] \\ &= \pi^2 + 5 = f(\pi) \end{aligned}

∴ f\therefore\ f is continuous at x=πx = \pi⁠.

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

The one idea this problem is built on

ff is built from three functions we already know are continuous everywhere: the polynomial x2x^2⁠, the sine function and the constant 55⁠. Theorem 1 says sums and differences of functions continuous at a point are continuous at that point. So the answer is "yes" without any special work at π\pi⁠. We also confirm it directly from the definition, so the reason is visible and not just quoted.

Given, to find, and the plan
Given
f(x)=x2−sin⁡x+5f(x)=x^2-\sin x+5 for real xx (with xx in radians).
To find
Whether lim⁡x→πf(x)=f(π)\lim_{x\to\pi}f(x)=f(\pi)⁠.
Plan
(1) Find f(π)f(\pi)⁠. (2) Name the continuous building blocks. (3) Apply Theorem 1. (4) Confirm with a direct limit, writing x=π+hx=\pi+h⁠.

Basics you need

1Continuity at a point

ff is continuous at cc when lim⁡x→cf(x)=f(c)\lim_{x\to c}f(x)=f(c)⁠, that is, both one-sided limits equal f(c)f(c) (Definition 1, Section 5.2). Basics: what a limit is

2Algebra of continuous functions

If ff and gg are continuous at cc⁠, then so are f+gf+g⁠, f−gf-g⁠, f⋅gf\cdot g⁠, and f/gf/g when g(c)≠0g(c)\ne0 (Theorem 1, Section 5.2.1). Basics: what a limit is

3Sine values and the angle-sum rule

sin⁡π=0\sin\pi=0 (radians: π\pi is 180∘180^\circ⁠). sin⁡(π+h)\sin(\pi+h) equals sin⁡πcos⁡h+cos⁡πsin⁡h\sin\pi\cos h+\cos\pi\sin h⁠, which is −sin⁡h-\sin h⁠. Basics: trigonometric identities

4sin⁡h→0\sin h\to0 as h→0h\to0

The fact the textbook uses in Example 17 to prove that sine is continuous. The book states it without proof (it says the fact is clear from the graph of sin⁡x\sin x near 00⁠). Basics: what a limit is

Solution

  1. π\pi is in the domain; find f(π)f(\pi)

    x2x^2⁠, sin⁡x\sin x and 55 are defined for every real xx⁠, so ff is defined on all of R\R⁠, including π\pi⁠. Since sin⁡π=0\sin\pi=0⁠,

    f(π)=π2−sin⁡π+5=π2−0+5=π2+5.\begin{aligned} f(\pi)&=\pi^2-\sin\pi+5\\ &=\pi^2-0+5=\pi^2+5. \end{aligned}
    Continuity at π\pi compares the limit with this value, so we need it first. Numerically π2+5≈14.869604\pi^2+5\approx14.869604⁠.
    sin⁡π=0\sin\pi=0⁠: the point at angle π\pi on the unit circle is (−1,0)(-1,0)⁠, whose yy-coordinate is 00⁠. Basics: trigonometric identities

  2. The three building blocks are continuous at π\pi

    x2x^2 is a polynomial, continuous at every real number (Example 14).

    sin⁡x\sin x is continuous at every real number (Example 17).

    The constant 55 is continuous at every real number (Example 5).

    Each of these was proved in the textbook for an arbitrary real number cc⁠, so each holds in particular at c=πc=\pi⁠.
    Polynomials, sine and constants are the basic continuous functions of Section 5.2. Lesson: continuity at a point

  3. Combine them with Theorem 1

    By Theorem 1 (2), x2−sin⁡xx^2-\sin x is continuous at π\pi (a difference of two functions continuous at π\pi⁠). By Theorem 1 (1), adding the constant function 55 keeps it continuous at π\pi⁠. So f(x)=x2−sin⁡x+5f(x)=x^2-\sin x+5 is continuous at x=πx=\pi.

    Theorem 1 is applied one operation at a time, exactly as ff is built: first the difference, then the sum. The theorem's condition, "both continuous at the same point cc", is met at each stage with c=πc=\pi⁠.
    Theorem 1 (1) and (2): sums and differences of continuous functions. Lesson: algebra of continuous functions

  4. Confirm directly from the definition

    Write x=π+hx=\pi+h⁠; then x→πx\to\pi means h→0h\to0⁠. Expanding the square and using sin⁡(π+h)=−sin⁡h\sin(\pi+h)=-\sin h⁠:

    f(π+h)=(π+h)2−sin⁡(π+h)+5=π2+2πh+h2+sin⁡h+5.\begin{aligned} f(\pi+h)&=(\pi+h)^2-\sin(\pi+h)+5\\ &=\pi^2+2\pi h+h^2+\sin h+5. \end{aligned}

    As h→0h\to0⁠: 2πh→02\pi h\to0⁠, h2→0h^2\to0 and sin⁡h→0\sin h\to0⁠. Hence

    lim⁡x→πf(x)=lim⁡h→0f(π+h)=π2+5=f(π).\begin{aligned} \lim_{x\to\pi}f(x)&=\lim_{h\to0}f(\pi+h)\\ &=\pi^2+5=f(\pi). \end{aligned}

    Nothing in this calculation depends on the sign of hh⁠, so the left-hand limit (h<0h\lt0⁠) and the right-hand limit (h>0h\gt0⁠) are both π2+5\pi^2+5⁠.

    This is the content of Theorem 1 made visible: each piece tends to its own value at π\pi⁠, so the whole expression does. The sign change in −sin⁡(π+h)=+sin⁡h-\sin(\pi+h)=+\sin h comes from cos⁡π=−1\cos\pi=-1 in the angle-sum formula.
    sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B\sin(A+B)=\sin A\cos B+\cos A\sin B⁠, with sin⁡π=0\sin\pi=0⁠, cos⁡π=−1\cos\pi=-1⁠. Basics: trigonometric identities
Result

Yes. lim⁡x→πf(x)=π2+5=f(π)\lim_{x\to\pi}f(x)=\pi^2+5=f(\pi)⁠, so ff is continuous at x=πx=\pi⁠. (In fact it is continuous at every real number, by the same argument.)

(Textbook answer: ff is continuous at x=πx=\pi⁠.)

Check it with numbers

f(π)=π2+5≈14.8696044f(\pi)=\pi^2+5\approx14.8696044⁠.

f(π−0.001)≈14.8623222f(\pi-0.001)\approx14.8623222 and f(π+0.001)≈14.8768886f(\pi+0.001)\approx14.8768886⁠. Both differ from f(π)f(\pi) by about 0.00730.0073⁠, which is close to (2π+1)(0.001)≈0.00728(2\pi+1)(0.001)\approx0.00728⁠, the first-order change predicted by 2πh+sin⁡h2\pi h+\sin h in Step 4. Halving the step halves the gap. ✓

Common mistakes
  1. Mixing degrees and radians. x=πx=\pi is the real number 3.14159…3.14159\ldots and sin⁡\sin is taken in radians. Treating π\pi as 180180 would make x2=32400x^2=32400⁠, which is meaningless here.
  2. Taking sin⁡π=1\sin\pi=1⁠. sin⁡π2=1\sin\frac{\pi}{2}=1⁠, but sin⁡π=0\sin\pi=0⁠.
  3. Quoting "it is continuous" without a reason. Name the building blocks and the parts of Theorem 1 used, or compute the limit.
  4. Sign error in sin⁡(π+h)\sin(\pi+h)⁠. It equals −sin⁡h-\sin h⁠, not sin⁡h\sin h⁠. (The limit is the same here, but the expansion in Step 4 would be wrong.)
What to take away
  1. For a function built by ++⁠, −-⁠, ×\times⁠, ÷\div from polynomials, sine, cosine and constants, Theorem 1 gives continuity at every point where no denominator is zero.
  2. A direct check with x=c+hx=c+h shows why: every piece tends to its own value.
  3. Always evaluate f(c)f(c) carefully; here f(π)=π2+5f(\pi)=\pi^2+5⁠.

Every fact used, and where it comes from

FactUsed inSource
Continuity at a pointSteps 3, 4Definition 1, Section 5.2 (Lesson: continuity at a point)
Constants, polynomials and sine are continuousStep 2Examples 5, 14 and 17, Section 5.2
Sum and difference of continuous functionsStep 3Theorem 1 (1), (2), Section 5.2.1 (Lesson: algebra of continuous functions)
sin⁡π=0\sin\pi=0⁠, cos⁡π=−1\cos\pi=-1⁠, angle-sum formulaSteps 1, 4Class XI, Chapter 3 (Trigonometric Functions)
lim⁡h→0sin⁡h=0\lim_{h\to0}\sin h=0Step 4Stated without proof in Example 17, Section 5.2.1
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