Continuity — Theory and Solved Exercises
Theoretical Recall
A function f is continuous at x₀ if it is defined at x₀ and:
\[\lim_{x\to x_0}f(x)=f(x_0).\]Equivalently, continuity at x₀ requires:
\[\lim_{x\to x_0^-}f(x) = \lim_{x\to x_0^+}f(x) = f(x_0).\]Epsilon–Delta Definition
A function f is continuous at x₀ if, for every ε > 0, there exists δ > 0 such that:
\[|x-x_0|<\delta \quad\Longrightarrow\quad |f(x)-f(x_0)|<\varepsilon.\]Types of Discontinuities
- Removable discontinuity: the limit exists and is finite, but f(x₀) is either undefined or different from the limit.
- Jump discontinuity: the left-hand and right-hand limits exist and are finite, but they are different.
- Infinite discontinuity: at least one of the one-sided limits is +∞ or −∞.
- Oscillatory discontinuity: the limit does not exist because the function oscillates indefinitely near the point.
Basic Continuity Properties
Polynomials are continuous on ℝ.
Rational functions are continuous wherever their denominators are nonzero.
The functions sin x, cos x and eˣ are continuous on ℝ.
The logarithmic function log x is continuous on its domain:
\[(0,+\infty).\]If f and g are continuous at x₀, then their sum, difference and product are continuous at x₀.
If additionally g(x₀) ≠ 0, their quotient is also continuous at x₀.
Compositions of continuous functions are continuous wherever the composition is defined.
Exercises
Exercise 1 — Removable Discontinuity
Consider:
\[f(x)=\frac{x^2-1}{x-1}, \qquad x\ne1,\]with:
\[f(1)=c.\]Determine c so that f is continuous at x = 1.
Solution.
For x ≠ 1, factor the numerator:
\[x^2-1=(x-1)(x+1).\]Therefore:
\[f(x)=x+1.\]Hence:
\[\lim_{x\to1}f(x) = \lim_{x\to1}(x+1) = 2.\]For continuity we require:
\[f(1) = \lim_{x\to1}f(x).\]Since f(1) = c:
\[c=2.\]Final Result
\[c=2\]Exercise 2 — Jump Discontinuity
Classify the discontinuity of sgn(x) at x = 0.
Solution.
The left-hand limit is:
\[\lim_{x\to0^-}\operatorname{sgn}(x)=-1.\]The right-hand limit is:
\[\lim_{x\to0^+}\operatorname{sgn}(x)=1.\]Since the two one-sided limits are finite but different:
\[-1\ne1.\]Therefore the two-sided limit does not exist and the function has a jump discontinuity.
Final Result
\[\text{Jump discontinuity at }x=0\]Exercise 3 — Infinite Discontinuity
Study:
\[f(x)=\frac1x\]at x = 0.
Solution.
Consider the one-sided limits:
\[\lim_{x\to0^+}\frac1x=+\infty,\]while:
\[\lim_{x\to0^-}\frac1x=-\infty.\]Thus the function has an infinite discontinuity at x = 0.
The vertical line x = 0 is also a vertical asymptote.
Final Result
\[\text{Infinite discontinuity at }x=0\]Exercise 4 — Extending a Function Continuously
Consider:
\[f(x)=\frac{\sin x}{x}, \qquad x\ne0,\]with:
\[f(0)=1.\]Study the continuity of f at x = 0.
Solution.
The fundamental trigonometric limit gives:
\[\lim_{x\to0}\frac{\sin x}{x}=1.\]By definition:
\[f(0)=1.\]Therefore:
\[\lim_{x\to0}f(x)=f(0).\]Hence f is continuous at x = 0.
For x ≠ 0, the quotient is continuous because both sin x and x are continuous and the denominator is nonzero.
Therefore f is continuous on all of ℝ.
Final Result
\[f\text{ is continuous on }\mathbb{R}\]Exercise 5 — Continuity by the Squeeze Theorem
Consider:
\[f(x)=|x|\sin\left(\frac1x\right), \qquad x\ne0,\]with:
\[f(0)=0.\]Study continuity at x = 0.
Solution.
Since:
\[-1\le \sin\left(\frac1x\right) \le1,\]we obtain:
\[\left| |x|\sin\left(\frac1x\right) \right| \le|x|.\]As x → 0:
\[|x|\to0.\]Therefore, by the Squeeze Theorem:
\[\lim_{x\to0} |x|\sin\left(\frac1x\right) = 0.\]Since:
\[f(0)=0,\]we have:
\[\lim_{x\to0}f(x)=f(0).\]Final Result
\[f\text{ is continuous at }x=0\]Exercise 6 — Continuity of a Piecewise Function
Consider:
\[f(x)= \begin{cases} ax+b, & x<2,\\ x^2, & x\ge2. \end{cases}\]Determine a and b so that f is continuous at x = 2.
Solution.
The left-hand limit is:
\[\lim_{x\to2^-}f(x)=2a+b.\]Since the second branch applies at x = 2:
\[f(2)=2^2=4.\]The right-hand limit is:
\[\lim_{x\to2^+}f(x)=4.\]Continuity requires:
\[2a+b=4.\]Thus there are infinitely many pairs (a,b) satisfying the condition.
Equivalently:
\[b=4-2a.\]Final Result
\[2a+b=4\]Exercise 7 — Continuity and Domain
Study the continuity of:
\[f(x)=\log(x^2-4).\]Solution.
The logarithm requires a strictly positive argument:
\[x^2-4>0.\]Factorizing:
\[(x-2)(x+2)>0.\]Therefore:
\[x<-2\]or:
\[x>2.\]Hence the domain is:
\[(-\infty,-2)\cup(2,+\infty).\]The polynomial x² − 4 is continuous on ℝ, and the logarithm is continuous for positive arguments.
Therefore their composition is continuous throughout its domain.
Final Result
\[f\text{ is continuous on } (-\infty,-2)\cup(2,+\infty)\]Exercise 8 — Oscillatory Discontinuity
Consider:
\[f(x)= \begin{cases} \sin\left(\frac1x\right), & x\ne0,\\ 0, & x=0. \end{cases}\]Check continuity at x = 0.
Solution.
As x approaches zero, 1/x becomes arbitrarily large in absolute value and:
\[\sin\left(\frac1x\right)\]oscillates between −1 and 1.
For example, consider the sequences:
\[x_n= \frac{1}{\frac{\pi}{2}+2\pi n}.\]Then:
\[x_n\to0\]and:
\[\sin\left(\frac1{x_n}\right)=1.\]Now consider:
\[y_n= \frac{1}{\frac{3\pi}{2}+2\pi n}.\]Then:
\[y_n\to0\]and:
\[\sin\left(\frac1{y_n}\right)=-1.\]The function therefore approaches different values along two sequences converging to zero.
Hence:
\[\lim_{x\to0}\sin\left(\frac1x\right)\]does not exist.
Thus f is not continuous at zero.
Final Result
\[\text{Oscillatory discontinuity at }x=0\]Exercise 9 — Piecewise Jump Discontinuity
Consider:
\[f(x)= \begin{cases} x^2, & x\le0,\\ 1, & x>0. \end{cases}\]Check continuity at x = 0.
Solution.
The left-hand limit is:
\[\lim_{x\to0^-}x^2=0.\]The right-hand limit is:
\[\lim_{x\to0^+}1=1.\]Thus:
\[\lim_{x\to0^-}f(x) \ne \lim_{x\to0^+}f(x).\]The two-sided limit does not exist.
Moreover:
\[f(0)=0.\]The value at zero agrees with the left-hand limit, but this does not restore continuity because the right-hand limit is different.
Final Result
\[\text{Jump discontinuity at }x=0\]Exercise 10 — Parameters in a Piecewise Function
Consider:
\[f(x)= \begin{cases} \cos x, & x<0,\\ a+bx, & x\ge0. \end{cases}\]Determine a and b so that f is continuous at x = 0.
Solution.
The left-hand limit is:
\[\lim_{x\to0^-}\cos x=1.\]For the second branch:
\[f(0)=a.\]The right-hand limit is:
\[\lim_{x\to0^+}(a+bx)=a.\]Continuity therefore requires:
\[a=1.\]The term bx tends to zero independently of the value of b, so continuity imposes no restriction on b.
Thus:
\[b\in\mathbb{R}.\]Final Result
\[a=1, \qquad b\in\mathbb{R}\]