Continuity and Continuous Functions

Problem
Check f(x)=2x+3 at x=1 and g(x)=x² at x=0.

1. 1.For the first function, the assigned value is f(1)=2·1+3=5.
2. 2.The limit is lim(x→1)(2x+3)=2·1+3=5, using the limit rules for sums and constant multiples. It equals the assigned value, so f is continuous at 1.
3. 3.For the second function, g(0)=0²=0. Also lim(x→0)x²=0²=0.
4. 4.The limit equals g(0), so g is continuous at zero. In both calculations we compared a limiting value with a separately identified function value.

Problem
Check the continuity of f(x)=|x| at zero.

1. 1.The function value is f(0)=0. Write |x|=−x when x<0 and |x|=x when x≥0.
2. 2.Approaching from the left gives lim(x→0⁻)(−x)=0. Approaching from the right gives lim(x→0⁺)x=0.
3. 3.Both limits equal 0, which is also f(0). Therefore |x| is continuous at zero.
4. 4.A change in the formula at a point does not by itself mean a discontinuity. The outputs, rather than the appearance of the formulas, decide the result.

Problem
Let f(x)=x³+3 for x≠0 and f(0)=1. Is f continuous at zero?

1. 1.The special definition gives f(0)=1; substituting zero in x³+3 would use the wrong branch for the assigned value.
2. 2.For nonzero x near zero the polynomial branch applies, so lim(x→0)f(x)=0³+3=3.
3. 3.The limit 3 differs from the function value 1. Hence f is discontinuous at zero. Replacing f(0) by 3 would repair this particular mismatch.

Problem
Show that a constant function f(x)=k and the identity function g(x)=x are continuous on the real line.

1. 1.Choose any real input c. For the constant function, f(c)=k and every nearby output is also k.
2. 2.Thus lim(x→c)f(x)=k=f(c). Since c was arbitrary, the constant function is continuous everywhere.
3. 3.For the identity function, g(c)=c and lim(x→c)x=c.
4. 4.Again the limit equals the value for every real c. The identity function is therefore continuous everywhere as well.

Problem
Explain why |x|, x³+x²−1, and a general polynomial are continuous functions.

1. 1.For |x|, the previous example handled c=0. If c<0, the nearby formula is −x, whose limit is −c=|c|. If c>0, the nearby formula is x, whose limit is c=|c|. These cases cover the real line.
2. 2.For p(x)=x³+x²−1, the limit rules give lim(x→c)p(x)=c³+c²−1=p(c) for every real c.
3. 3.The same reasoning applies to p(x)=a₀+a₁x+⋯+aₙxⁿ: each power tends to the corresponding power of c, and a finite sum of these terms tends to p(c).
4. 4.Thus every polynomial is continuous on the real line. This conclusion follows from limit laws; it is not an assumption based only on a smooth-looking drawing.

Problem
Discuss f(x)=1/x, with domain x≠0.

1. 1.Choose any nonzero real number c. The quotient limit rule applies because its denominator tends to the nonzero number c.
2. 2.It gives lim(x→c)1/x=1/c. This is exactly f(c).
3. 3.Every point of the domain satisfies continuity. Therefore the reciprocal is a continuous function on its domain, even though zero is excluded.

Problem
Compare f(x)=x+2 for x≤1, f(x)=x−2 for x>1, with g(x)=x+2 for x<1, g(1)=0, g(x)=x−2 for x>1.

1. 1.For either function, a point c<1 has a neighbourhood using only x+2, so continuity holds there. At c>1 the formula x−2 similarly gives continuity.
2. 2.At 1 the left limit is 1+2=3 and the right limit is 1−2=−1. The limits disagree.
3. 3.The first function assigns f(1)=3; the second assigns g(1)=0. Neither assignment changes the two approaching values.
4. 4.Both functions are discontinuous only at 1. No single replacement value at 1 can remove a jump with unequal one-sided limits.

Problem
Let f(x)=x+2 for x<0 and f(x)=−x+2 for x>0. Is the function continuous?

1. 1.Its domain is all real numbers except zero. Every negative input has a neighbourhood on which f is the polynomial x+2.
2. 2.Every positive input has a neighbourhood on which f is the polynomial −x+2. Continuity therefore holds at every domain point.
3. 3.The given function is continuous on its domain. Both branches approach 2 at zero, so an extension defined by f(0)=2 would also be continuous on the whole real line.
4. 4.The extension is a new definition. We must not silently act as though the original function already had a value at zero.

Problem
Let f(x)=x² for x<0 and f(x)=x for x≥0. Discuss continuity.

1. 1.Away from zero, the relevant polynomial formula is continuous. Only zero needs a separate calculation.
2. 2.At zero the second branch gives f(0)=0. The left limit of x² is 0 and the right limit of x is 0.
3. 3.The one-sided limits agree with f(0). Hence the function is continuous at zero and at every other real input.
4. 4.The change from a parabola to a line does not break continuity. Whether the slopes match will be a separate question in the next lesson.

Problem
Show where f(x)=[x] is continuous and where it is discontinuous.

1. 1.If c is not an integer, choose inputs close enough to c that no integer lies between them and c. Their greatest integer values are all [c]. Thus the limit is [c]=f(c).
2. 2.If c=n is an integer, inputs just below n have value n−1, whereas inputs just above n have value n.
3. 3.The left limit is n−1 and the right limit is n=f(n). The unequal limits imply a discontinuity at n.
4. 4.Therefore the points of discontinuity are precisely all integers. The related function x−[x] has left limit 1 and right limit 0 at each integer, so it also has a jump at every integer.

Problem
Explain why every rational function p(x)/q(x) is continuous wherever it is defined.

1. 1.The numerator p and denominator q are polynomials, so they are continuous at every real input.
2. 2.The domain of their quotient excludes the zeros of q. At any remaining input c, q(c)≠0, so the quotient continuity rule applies.
3. 3.For example, (x²−25)/(x+5) equals x−5 for x≠−5. This makes its continuity on the original domain easy to see, but cancellation does not add −5 to that domain.
4. 4.The limit at −5 is −10. We could create a continuous extension by assigning that value, but the uncancelled rational function itself excludes the input.

Problem
Show that sin(x²) and |1−x+|x|| are continuous functions.

1. 1.The inner function x² is continuous everywhere. Sine is continuous at every possible output x². Their composition sin(x²) is therefore continuous everywhere.
2. 2.For the second function, |x| is continuous and so is the polynomial 1−x. Their sum g(x)=1−x+|x| is continuous.
3. 3.Applying the continuous outer modulus function to g gives |g(x)|. Thus |1−x+|x|| is continuous everywhere.
4. 4.The same reasoning establishes continuity of cos(x²), |cos x|, and sin|x| without lengthy piecewise expansions.

Problem
First let f(x)=ax+1 for x≤3 and bx+3 for x>3. Then consider g(x)=5 for x≤2, ax+b for 2<x<10, and 21 for x≥10.

1. 1.For f at 3, the left limit and value are 3a+1, while the right limit is 3b+3. Continuity requires 3a+1=3b+3, or a−b=2/3.
2. 2.For g at 2, the middle branch must approach 5: 2a+b=5. At 10, it must approach 21: 10a+b=21.
3. 3.Subtracting the equations gives 8a=16, hence a=2. Substitution gives b=1.
4. 4.With these values the middle segment is 2x+1. Its values match both constant pieces at their joins, and each piece is already continuous away from those joins.

Problem
Let f(x)=k cos x/(π−2x) for x≠π/2 and f(π/2)=3. Find k for continuity at π/2.

1. 1.At the joining input the assigned value is 3. The quotient gives 0/0 on direct substitution, so substitution alone does not find its limit.
2. 2.Put x=π/2+h. Then cos x=−sin h and π−2x=−2h, so the quotient becomes (k/2)(sin h/h) for h≠0.
3. 3.The known trigonometric limit sin h/h→1 gives a two-sided limit k/2.
4. 4.Continuity requires k/2=3, hence k=6. Both approaches use the same transformed limit, which then matches the assigned value.

Problem
Let f(x)=λ(x²−2x) for x≤0 and f(x)=4x+1 for x>0. Can any λ make it continuous at zero? What happens at 1?

1. 1.The value at zero and the left limit are λ·0=0 for every λ.
2. 2.The right limit at zero is 4·0+1=1. Because the parameter never changes the zero on the left, no choice of λ can make the two limits agree.
3. 3.At 1, nearby inputs all use 4x+1. Its limit is 5=f(1), so the function is continuous at 1 for every λ.
4. 4.A discontinuity at one input does not prevent continuity at another input.

Problem
Define f(x)=x²sin(1/x) for x≠0 and f(0)=0. Discuss its continuity.

1. 1.For x≠0, reciprocal, sine, and multiplication are continuous on the relevant domains, so f is continuous away from zero.
2. 2.Although sin(1/x) oscillates as x→0, its value always lies between −1 and 1. Multiplying by x² gives −x²≤f(x)≤x².
3. 3.Both bounding expressions approach 0. The squeezed function therefore also approaches 0.
4. 4.This equals f(0), so the whole function is continuous. Oscillation alone is not decisive; the shrinking multiplier controls its size.