Chapter 5: Continuity and Differentiability

Given: $f(x) = 5x – 3$

Continuity at $x = 0$: Evaluate the limit by direct substitution:

Putting $x = 0$: $= 5(0) – 3 = -3$

Also, $f(0) = 5(0) – 3 = -3$

Continuity at $x = -3$:

Putting $x = -3$: $= 5(-3) – 3 = -18$

Also, $f(-3) = 5(-3) – 3 = -18$

Continuity at $x = 5$:

Putting $x = 5$: $= 5(5) – 3 = 22$

Also, $f(5) = 5(5) – 3 = 22$

Given: $f(x) = 2x^2 – 1$

Compute the limit at $x = 3$ by direct substitution:

Putting $x = 3$: $= 2(3)^2 – 1 = 18 – 1 = 17$

Also, $f(3) = 2(3)^2 – 1 = 17$

(a) The domain of $f$ is $\mathbf{R}$. Let $c$ be any real number:

Since the limit equals the function value at every point, $f$ is continuous on $\mathbf{R}$.

Alternatively: $f(x) = x – 5$ is a polynomial, and every polynomial is continuous on its domain.

(b) The domain of $f$ is $\mathbf{R} – \{5\}$ (since $f(5) \to \infty$). For any $c \neq 5$:

$f$ is continuous at every point of its domain. Alternatively: $f$ is a rational function with non-zero denominator on its domain, hence continuous.

(c) The domain is $\mathbf{R} – \{-5\}$. Note that

This reduces to a polynomial on its domain, so $f$ is continuous on $\mathbf{R} – \{-5\}$.

(d) $f(x) = |x – 5|$ is a modulus function. Every modulus function is continuous on $\mathbf{R}$, so $f$ is continuous on $\mathbf{R}$.

Since $f(x) = x^n$ is a polynomial (for positive integer $n$), it is continuous everywhere on $\mathbf{R}$, and hence continuous at $x = n$ in particular.

Direct verification:

Also, $f(n) = n^n$.

Continuity at $x = 0$:

Since $x \to 0^-$ and $x \to 0^+$ both give values less than 1, both one-sided limits use $f(x) = x$:

Continuity at $x = 1$:

$\therefore f(x)$ is discontinuous at $x = 1$.

Continuity at $x = 2$:

For $x$ near 2 (both sides), $x > 1$ so $f(x) = 5$:

Answer: $f$ is continuous at $x = 0$ and $x = 2$, but not at $x = 1$.

For $x < 2$, $f(x) = 2x+3$ is a polynomial, hence continuous. For $x > 2$, $f(x) = 2x-3$ is a polynomial, hence continuous. The only point to check is the partition point $x = 2$:

The limit does not exist at $x = 2$, so $f$ is discontinuous at $x = 2$ only.

The domain is $\mathbf{R}$. Each piece is a polynomial/absolute value, hence continuous within its interval. We check the two partition points.

At $x = -3$:

Both one-sided limits equal 6 and $f(-3) = |-3|+3 = 6$. $\therefore f$ is continuous at $x = -3$.

At $x = 3$:

$f$ is discontinuous at $x = 3$ only.

Notice that $\dfrac{|x|}{x} = 1$ for $x > 0$ and $= -1$ for $x < 0$, so $f$ behaves like a constant on each half-line and is continuous away from $x = 0$.

Checking the partition point $x = 0$:

The limit does not exist at $x = 0$. $\therefore f$ is discontinuous at $x = 0$ only.

For $x < 0$: $\dfrac{x}{|x|} = \dfrac{x}{-x} = -1$. So in fact $f(x) = -1$ for all real $x$.

Since $f(x) = -1$ is a constant function, it is continuous everywhere on $\mathbf{R}$. There are no points of discontinuity.

Both pieces are polynomials, hence continuous away from the partition point $x = 1$. Check $x = 1$:

$\therefore f$ is continuous on $\mathbf{R}$. There are no points of discontinuity.

Both pieces are polynomials. Check partition point $x = 2$:

There are no points of discontinuity.

Both pieces are polynomials. Check partition point $x = 1$:

The limit does not exist. $\therefore f$ is discontinuous at $x = 1$ only.

Both pieces are polynomials, continuous on their respective intervals. Check the partition point $x = 1$:

The limit does not exist at $x = 1$. $\therefore f$ is not a continuous function; $x = 1$ is its only point of discontinuity.

Each piece is a constant function, so $f$ is continuous within each interval. The partition points are $x = 1$ and $x = 3$.

At $x = 1$:

Limits are unequal $\Rightarrow$ $f$ is discontinuous at $x = 1$.

At $x = 3$:

Limits are unequal $\Rightarrow$ $f$ is discontinuous at $x = 3$.

$\therefore$ The two points of discontinuity in $[0,10]$ are $x = 1$ and $x = 3$.

Each piece is a polynomial/constant, continuous within its interval. Check the partition points.

At $x = 0$:

At $x = 1$:

Left and right limits are unequal $\Rightarrow$ $f$ is discontinuous at $x = 1$ only.

Check the two partition points $x = -1$ and $x = 1$.

At $x = -1$:

At $x = 1$:

$\therefore f$ is continuous on its entire domain $\mathbf{R}$.

For continuity at $x = 3$, the left limit, right limit and function value must all be equal:

Setting left limit $=$ right limit:

Continuity at $x = 0$:

Since $0 \neq 1$ regardless of $\lambda$, the limit does not exist at $x = 0$. Therefore, for no value of $\lambda$ is $f$ continuous at $x = 0$.

Continuity at $x = 1$:

For $x$ near 1 (both sides), $x > 0$ so $f(x) = 4x+1$:

$\therefore f$ is continuous at $x = 1$ for all real values of $\lambda$.

Let $c$ be any integer. Substitute $x = c – h$ (with $h \to 0^+$) for the left limit:

Since $c \in \mathbb{Z}$ and $h \to 0^+$, we have $[c-h] = c-1$:

For the right limit, substitute $x = c + h$:

The limit fails to exist at every integer $c$. $\therefore g(x) = x – [x]$ is discontinuous at all integral points.

Write $f(x) = g(x) – h(x)$ where $g(x) = x^2+5$ (polynomial, continuous) and $h(x) = \sin x$ (sine function, continuous). Their difference is continuous everywhere, so $f$ is continuous at $x = \pi$.

Direct verification:

Both $\sin x$ and $\cos x$ are continuous for all real $x$. By the algebra of continuous functions:

(a) Their sum $f(x) = \sin x + \cos x$ is continuous for all $x \in \mathbf{R}$.

(b) Their difference $f(x) = \sin x – \cos x$ is continuous for all $x \in \mathbf{R}$.

(c) Their product $f(x) = \sin x \cdot \cos x$ is continuous for all $x \in \mathbf{R}$.

(i) $f(x) = \cos x$: Domain is $\mathbf{R}$. For any $c \in \mathbf{R}$, put $x = c+h$, $h \to 0$:

Since the limit equals $f(c) = \cos c$, $\cos x$ is continuous on $\mathbf{R}$.

(ii) $f(x) = \operatorname{cosec} x = \dfrac{1}{\sin x}$: Domain is $\mathbf{D} = \mathbf{R} – \{x = n\pi;\ n \in \mathbb{Z}\}$. On this domain, $\sin x \neq 0$ and is continuous, so the quotient $\dfrac{1}{\sin x}$ is continuous on $\mathbf{D}$.

(iii) $f(x) = \sec x = \dfrac{1}{\cos x}$: Domain is $\mathbf{D} = \mathbf{R} – \left\{x = (2n+1)\dfrac{\pi}{2};\ n \in \mathbb{Z}\right\}$. On $\mathbf{D}$, $\cos x \neq 0$, so $\sec x$ is continuous on $\mathbf{D}$.

(iv) $f(x) = \cot x = \dfrac{\cos x}{\sin x}$: Domain is $\mathbf{D} = \mathbf{R} – \{x = n\pi;\ n \in \mathbb{Z}\}$. Both numerator and denominator are continuous, and denominator is non-zero on $\mathbf{D}$, so $\cot x$ is continuous on $\mathbf{D}$.

For $x < 0$, $\dfrac{\sin x}{x}$ is a ratio of continuous functions with $x \neq 0$, so $f$ is continuous for $x < 0$. For $x > 0$, $f(x) = x+1$ is a polynomial, so $f$ is continuous for $x > 0$. Check $x = 0$:

$f$ is continuous at $x = 0$ as well. $\therefore f$ is continuous on $\mathbf{R}$ — no points of discontinuity.

For $x \neq 0$, $f(x) = x^2 \sin\dfrac{1}{x}$ is a product of the continuous function $x^2$ and the bounded function $\sin\dfrac{1}{x}$, so $f$ is continuous for all $x \neq 0$.

At $x = 0$:

$f$ is continuous at $x = 0$ as well. $\therefore f$ is continuous on $\mathbf{R}$.

For $x \neq 0$, $f(x) = \sin x – \cos x$ is continuous (difference of two continuous functions). At $x = 0$:

$f$ is continuous at $x = 0$ as well. $\therefore f$ is continuous on $\mathbf{R}$.

Compute the limit by substituting $x = \dfrac{\pi}{2} – h$, $h \to 0^+$:

(The same result is obtained from the right side.) For continuity:

For continuity: $4k = 3$

For continuity: $k\pi + 1 = -1$

For continuity: $5k+1 = 10$

At $x = 2$: continuity requires

At $x = 10$: continuity requires

Subtract (v) from (vii): $8a = 16 \Rightarrow a = 2$. Substituting back: $4+b = 5 \Rightarrow b = 1$.

Write $f = g \circ h$ where $g(x) = \cos x$ (cosine — continuous) and $h(x) = x^2$ (polynomial — continuous).

Since the composite of two continuous functions is continuous, $f(x) = \cos(x^2)$ is continuous on $\mathbf{R}$.

Write $f = h \circ g$ where $g(x) = \cos x$ (continuous) and $h(x) = |x|$ (modulus — continuous).

Since the composite of two continuous functions is continuous, $f(x) = |\cos x|$ is continuous on $\mathbf{R}$.

Let $f(x) = \sin x$ (continuous) and $g(x) = |x|$ (continuous). Then

The composite of two continuous functions is continuous, so $\sin|x|$ is continuous on $\mathbf{R}$.

The critical points where the modulus expressions change sign are $x = 0$ and $x = -1$. Express $f$ piece-wise on the three intervals $(-\infty,-1]$, $[-1,0]$, $[0,\infty)$:

For $x \leq -1$: $|x| = -x$, $|x+1| = -(x+1)$, so $f(x) = -x+(x+1) = 1$.

For $-1 \leq x \leq 0$: $|x| = -x$, $|x+1| = x+1$, so $f(x) = -x-(x+1) = -2x-1$.

For $x \geq 0$: $|x| = x$, $|x+1| = x+1$, so $f(x) = x-(x+1) = -1$.

At $x = -1$:

Continuous at $x = -1$.

At $x = 0$:

Continuous at $x = 0$.

$\therefore f$ is continuous on all of $\mathbf{R}$. There are no points of discontinuity.

Alternatively: $|x|$ and $|x+1|$ are both continuous on $\mathbf{R}$, so their difference is continuous on $\mathbf{R}$.

Apply the chain rule — differentiate the outer sine function, then multiply by the derivative of the inner expression:

Apply the chain rule three times — sec, then tan, then $\sqrt{x}$:

Apply the quotient rule, then the chain rule for each trig function:

Use the product rule on $u = \cos x^3$ and $v = \sin^2(x^5)$:

Write $y = 2(\cot x^2)^{1/2}$ and apply the chain rule:

At $x = 1$, $f(1) = 0$. Compute the one-sided derivatives:

At $x = 1$: $f(1) = [1] = 1$. Substitute $x = 1-h$, $h \to 0^+$:

The left-hand derivative does not exist $\Rightarrow$ $f$ is not differentiable at $x = 1$.

At $x = 2$: $f(2) = [2] = 2$. Substitute $x = 2-h$, $h \to 0^+$:

The left-hand derivative does not exist $\Rightarrow$ $f$ is not differentiable at $x = 2$.

Differentiate both sides w.r.t. $x$:

Differentiate both sides w.r.t. $x$:

Differentiate both sides, using the product rule on $xy$:

Substitute $x = \tan\theta$ to simplify the inverse trig expression:

Substitute $x = \tan\theta$:

Substitute $x = \tan\theta$:

Substitute $x = \tan\theta$:

Substitute $x = \tan\theta$:

Substitute $x = \sin\theta$:

Substitute $x = \cos\theta$:

Apply the quotient rule:

Apply the chain rule for sin, then $\tan^{-1}$, then $e^{-x}$:

Differentiate each term separately using the chain rule:

Write $y = (e^{\sqrt{x}})^{1/2}$ and apply chain rule:

Take logarithm of both sides to convert the product to a sum:

Differentiate both sides w.r.t. $x$:

Take logarithm to separate the factors:

Differentiate:

Take log and use the product rule:

Set $u = x^x$ and $v = 2^{\sin x}$, so $\dfrac{dy}{dx} = \dfrac{du}{dx} – \dfrac{dv}{dx}$.

For $u = x^x$: $\log u = x\log x \Rightarrow \dfrac{du}{dx} = x^x(1+\log x)$

For $v = 2^{\sin x}$: $\dfrac{dv}{dx} = 2^{\sin x}(\log 2)\cos x$

Take logarithm to split the product into a sum:

Set $u = \left(x+\frac{1}{x}\right)^x$ and $v = x^{(1+1/x)}$.

For $u$: $\log u = x\log\!\left(x+\frac{1}{x}\right)$, differentiating:

For $v$: $\log v = \left(1+\frac{1}{x}\right)\log x$, differentiating:

Set $u = (\log x)^x$ and $v = x^{\log x}$.

For $u$: $\log u = x\log(\log x)$, so

For $v$: $\log v = (\log x)^2$, so

Set $u = (\sin x)^x$ and $v = \sin^{-1}\sqrt{x}$.

For $u$: $\log u = x\log\sin x$, so

For $v$:

Set $u = x^{\sin x}$ and $v = (\sin x)^{\cos x}$.

Set $u = x^{x\cos x}$ and $v = \dfrac{x^2+1}{x^2-1}$.

For $u$: $\log u = x\cos x\log x$, differentiating with the product rule on three factors:

For $v$ (quotient rule):

Set $u = (x\cos x)^x$ and $v = (x\sin x)^{1/x}$.

For $u$: $\log u = x(\log x+\log\cos x)$, so

For $v$: $\log v = \dfrac{1}{x}(\log x+\log\sin x)$, so

Set $u = x^y$ and $v = y^x$, so $u + v = 1$ and $\dfrac{du}{dx} + \dfrac{dv}{dx} = 0$.

Substitute into $\dfrac{du}{dx} + \dfrac{dv}{dx} = 0$ and solve for $\dfrac{dy}{dx}$:

Taking logarithms: $y\log x = x\log y$. Differentiate both sides:

Taking logarithms: $y\log\cos x = x\log\cos y$. Apply the product rule on each side:

Take logarithms: $\log x + \log y = x – y$. Differentiate both sides:

Take logarithm to convert the product to a sum, then differentiate:

At $x = 1$: $f(1) = 2\cdot2\cdot2\cdot2 = 16$ and each term in the bracket equals $\frac{1}{2}$:

(i) Product rule:

(ii) Expand first: $y = x^5-5x^4+15x^3-26x^2+11x+72$

(iii) Logarithmic differentiation:

All three methods yield the same answer: $\dfrac{dy}{dx} = 5x^4-20x^3+45x^2-52x+11$.

(i) Repeated product rule: Treat $uv$ as one factor:

(ii) Logarithmic differentiation: Let $y = uvw$, so $\log y = \log u+\log v+\log w$:

After applying the quotient rule to both $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$, simplifying, and dividing:

Write $x = a^{\frac{1}{2}\sin^{-1}t}$ and $y = a^{\frac{1}{2}\cos^{-1}t}$. Differentiating w.r.t. $t$:

Since $y = \cos^{-1}x$, we have $x = \cos y$.

Differentiate and multiply by $x$ to clear the denominator:

Differentiate again using the product rule on the left, and the chain rule on the right:

Multiply through by $x$:

Substitute into the LHS:

(All terms cancel pairwise.)

From $e^y(x+1) = 1$: $y = -\log(x+1)$.

Differentiate again:

Multiply both sides by $(1+x^2)$:

Take logarithm: $\log y = 3\cos 2x\log(5x)$. Differentiate:

Note $x\sqrt{x} = x^{3/2}$:

Apply the quotient rule:

Simplify using half-angle identities: for $0 < x < \frac{\pi}{2}$,

The fraction simplifies to $\cot\dfrac{x}{2}$, so

Take log: $\log y = \log x\cdot\log(\log x)$. Differentiate:

Take log: $\log y = (\sin x-\cos x)\log(\sin x-\cos x)$. Differentiate using the product rule:

Differentiate term by term. Note $a^a$ is a constant with zero derivative. For $u = x^x$: $\log u = x\log x$, giving $\dfrac{du}{dx} = x^x(1+\log x)$.

Set $u = x^{x^2-3}$ and $v = (x-3)^{x^2}$.

For $u$: $\log u = (x^2-3)\log x$, so $\dfrac{du}{dx} = x^{x^2-3}\!\left(\dfrac{x^2-3}{x}+2x\log x\right)$

For $v$: $\log v = x^2\log(x-3)$, so $\dfrac{dv}{dx} = (x-3)^{x^2}\!\left(\dfrac{x^2}{x-3}+2x\log(x-3)\right)$

Rearrange and square to find $y$ explicitly. From $x\sqrt{1+y} = -y\sqrt{1+x}$, squaring gives $x^2(1+y) = y^2(1+x)$, which factors as $(x-y)(x+y) = -xy(x-y)$. Dividing by $(x-y) \neq 0$:

Differentiate:

Differentiate: $\dfrac{dy}{dx} = -\dfrac{x-a}{y-b}$

Differentiate again and simplify using $(x-a)^2+(y-b)^2 = c^2$:

Substitute:

This is a constant independent of $a$ and $b$. $\blacksquare$

Express $x = \dfrac{\cos y}{\cos(\alpha+y)}$ and differentiate w.r.t. $y$ using the quotient rule:

Take the reciprocal:

Differentiate again w.r.t. $x$:

Express piece-wise: $f(x) = x^3$ for $x \geq 0$ and $f(x) = -x^3$ for $x < 0$.

Then $f'(x) = 3x^2$ for $x > 0$ and $f'(x) = -3x^2$ for $x < 0$.

At $x = 0$: $Lf'(0) = Rf'(0) = 0$, so $f'(0) = 0$.

Similarly, $Lf”(0) = Rf”(0) = 0$, confirming $f”(0) = 0$.

Base case $P(1)$: $\dfrac{d}{dx}(x^1) = 1 = 1\cdot x^0$. ✓

Inductive step: Assume $P(k)$ holds: $\dfrac{d}{dx}(x^k) = kx^{k-1}$. Then:

This is exactly $P(k+1)$. By the principle of mathematical induction, the result holds for all positive integers $n$. $\blacksquare$

Treat $A$ and $B$ as functions of $x$ and differentiate both sides:

Dividing both sides by $\dfrac{dA}{dx}+\dfrac{dB}{dx}$:

Yes. A concrete example is $f(x) = |x-1|+|x-2|$.

This simplifies piece-wise to: $f(x) = 3-2x$ for $x \leq 1$; $f(x) = 1$ for $1 \leq x \leq 2$; $f(x) = 2x-3$ for $x \geq 2$.

Checking at $x = 1$: both one-sided limits equal 1, and $f(1) = 1$, so $f$ is continuous there. However, the left derivative is $-2$ and the right derivative is $0$, so $f$ is not differentiable at $x = 1$.

Checking at $x = 2$: both one-sided limits equal 1 and $f(2) = 1$, so $f$ is continuous there. The left derivative is $0$ and the right derivative is $2$, so $f$ is not differentiable at $x = 2$.

$\therefore f(x) = |x-1|+|x-2|$ is continuous on $\mathbf{R}$ but fails to be differentiable at exactly two points.

Expand the determinant along the first row:

Differentiate w.r.t. $x$:

This is precisely the expansion of $\begin{vmatrix} f'(x) & g'(x) & h'(x) \\ l & m & n \\ a & b & c \end{vmatrix}$. $\blacksquare$

Differentiate both sides w.r.t. $x$:

Multiply through by $\sqrt{1-x^2}$: