Continuity and Differentiability

sources:

• text: Standard textbook reference

Continuity ensures a function has no breaks, jumps, or holes. Differentiability ensures a function has a defined derivative. Every differentiable function is continuous, but not vice versa.

• $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$
• Left-hand limit: $\lim_{x \to a^-} f(x)$, right-hand limit: $\lim_{x \to a^+} f(x)$
• $f$ is differentiable at $x = a$ if $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$ exists
• Chain rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$
• Implicit differentiation: differentiate both sides with respect to $x$
• Logarithmic differentiation: take $\ln$ of both sides before differentiating

Problem: Determine whether $f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & x \neq 2 \\ 6 & x = 2 \end{cases}$ is continuous at $x = 2$.

Solution:

For $x \neq 2$: $f(x) = \frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2$

Left-hand limit: $\lim_{x \to 2^-} f(x) = 2 + 2 = 4$

Right-hand limit: $\lim_{x \to 2^+} f(x) = 2 + 2 = 4$

But $f(2) = 6 \neq 4$.

Since $\lim_{x \to 2} f(x) \neq f(2)$, the function is not continuous at $x = 2$.

Common mistake: Evaluating the limit by direct substitution when the function has a removable discontinuity. Simplify first.

Problem: Find $\frac{d}{dx}[\sin(\ln(x^2 + 1))]$.

Solution:

Let $u = \ln(x^2 + 1)$, $v = \sin(u)$:

$\frac{d}{dx}[\sin(\ln(x^2 + 1))] = \cos(\ln(x^2 + 1)) \cdot \frac{1}{x^2 + 1} \cdot 2x$

$= \frac{2x \cos(\ln(x^2 + 1))}{x^2 + 1}$

Common mistake: Forgetting one of the chain rule steps. Each nested function requires one more derivative factor.

Problem: Find $\frac{dy}{dx}$ if $y = x^x$.

Solution:

Take $\ln$ of both sides: $\ln y = x \ln x$

Differentiate implicitly: $\frac{1}{y}\frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$

$\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$

Common mistake: Trying to use the power rule for $x^x$. The power rule applies to $x^n$ (constant exponent), not $a^x$ (constant base).

1. Determine if $f(x) = |x - 3|$ is differentiable at $x = 3$.
2. Find $\frac{dy}{dx}$ if $y = e^{\sin x}$.
3. Find $\frac{dy}{dx}$ if $y = (\sin x)^x$.

Continuity and differentiability are the foundations of calculus. They guarantee that functions behave well enough for differentiation and integration, which are used throughout physics, engineering, economics, and biology.

A function is continuous if you can draw it without lifting your pen: Continuity means there are no holes, jumps, or vertical asymptotes — the function flows smoothly. Think of it as a road with no potholes or cliffs. Differentiability is stricter: it means the road not only exists but has a well-defined slope at every point (no sharp corners). The absolute value function |x| is continuous at x = 0 (no hole) but not differentiable there (sharp V-shape). Chain rule, implicit differentiation, and logarithmic differentiation are just tools for finding slopes of curves that are hard to differentiate directly.

Why it matters: Continuity and differentiability are the foundations of calculus — they guarantee that limits, derivatives, and integrals work properly. Every physical law expressed as a differential equation assumes continuity. Without these concepts, we couldn’t model motion, growth, decay, or virtually any changing quantity.

The key insight: Differentiability implies continuity (if a function has a derivative, it must be continuous), but continuity does NOT imply differentiability (a continuous function can have sharp corners). This one-way relationship is a fundamental theorem of calculus.

• Check continuity by verifying left limit = right limit = function value
• Differentiability implies continuity, but continuity does not imply differentiability
• Chain rule problems often have 2-3 nested functions
• Logarithmic differentiation is useful for functions of the form $f(x)^{g(x)}$
• Piecewise functions often have discontinuities at the boundaries

• Continuity: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a)$
• Derivative definition: $f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$
• Chain rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
• Implicit differentiation: differentiate both sides, then solve for $\frac{dy}{dx}$
• Logarithmic differentiation: $\frac{d}{dx}[f(x)] = f(x) \cdot \frac{d}{dx}[\ln f(x)]$

Problem: Find the value of $k$ if $f(x) = \begin{cases} kx + 1 & x \leq 3 \\ 2x - 1 & x > 3 \end{cases}$ is continuous at $x = 3$.

Solution:

For continuity at $x = 3$, we need $\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$.

Left-hand limit: $\lim_{x \to 3^-} f(x) = k(3) + 1 = 3k + 1$

Right-hand limit: $\lim_{x \to 3^+} f(x) = 2(3) - 1 = 5$

Function value: $f(3) = 3k + 1$

Setting equal: $3k + 1 = 5 \implies k = \frac{4}{3}$

Common mistake: Forgetting to check that the function value equals the limits, not just that the left and right limits are equal.

Problem: Find $\frac{d}{dx} \left[ \ln\left(\sqrt{\sin(x^2)}\right) \right]$.

Solution:

Simplify first: $\ln\left(\sqrt{\sin(x^2)}\right) = \frac{1}{2} \ln(\sin(x^2))$

Now differentiate: $\frac{d}{dx} \left[ \frac{1}{2} \ln(\sin(x^2)) \right] = \frac{1}{2} \cdot \frac{1}{\sin(x^2)} \cdot \cos(x^2) \cdot 2x$

$= \frac{x \cos(x^2)}{\sin(x^2)} = x \cot(x^2)$

Common mistake: Not simplifying before differentiating. Simplifying $\ln(\sqrt{u})$ to $\frac{1}{2}\ln u$ makes the chain rule much easier.

Problem: Find $\frac{dy}{dx}$ if $x^2 + y^2 + \sin(xy) = 4$.

Solution:

Differentiate both sides with respect to $x$: $2x + 2y \frac{dy}{dx} + \cos(xy) \left(y + x \frac{dy}{dx}\right) = 0$

Expand: $2x + 2y \frac{dy}{dx} + y\cos(xy) + x\cos(xy) \frac{dy}{dx} = 0$

Collect $\frac{dy}{dx}$ terms: $\frac{dy}{dx} \left(2y + x\cos(xy)\right) = -2x - y\cos(xy)$

$\frac{dy}{dx} = \frac{-2x - y\cos(xy)}{2y + x\cos(xy)}$

Common mistake: Forgetting to apply the chain rule when differentiating $\sin(xy)$. The derivative of $\sin(xy)$ is $\cos(xy) \cdot (y + x \frac{dy}{dx})$.

1. For piecewise functions, always check continuity at the boundary by computing left and right limits separately
2. The function $|x - a|$ is continuous everywhere but not differentiable at $x = a$
3. When using logarithmic differentiation, remember that $\ln(ab) = \ln a + \ln b$ can simplify products
4. For implicit differentiation, always collect $\frac{dy}{dx}$ terms on one side before solving
5. The derivative of $a^x$ is $a^x \ln a$ (not $xa^{x-1}$), while the derivative of $x^n$ is $nx^{n-1}$ (not $n^x \ln x$)

Confusing continuity with differentiability. A function can be continuous but not differentiable (like |x| at x = 0). Continuity means no breaks or jumps; differentiability means the derivative exists. Students often assume continuity implies differentiability, which is false.

Forgetting that the derivative of a^x is a^x ln(a), not x*a^(x-1). The power rule x^n -> nx^(n-1) applies only when x is the variable and n is constant. For a^x where a is constant, the derivative is a^x ln(a). Students frequently apply the power rule to exponential functions.

Making sign errors in implicit differentiation. When differentiating y^2 implicitly, the result is 2y * dy/dx, not just 2y. Students often forget the chain rule factor dy/dx when differentiating terms involving y. Always multiply by dy/dx when differentiating any term containing y with respect to x.

• Derivatives — Differentiability implies continuity, making continuity a prerequisite for understanding derivatives.
• Integrals — The definite integral requires the function to be continuous over the interval of integration.
• Limits — Continuity is defined in terms of limits, connecting the concept to the foundational notion of limiting behaviour.