Continuity Questions in English

(B) For $x > 0$, $|x| = x$, so $f(x) = \frac{x - x}{x} = 0$.
For $x < 0$, $|x| = -x$, so $f(x) = \frac{x - (-x)}{x} = \frac{2x}{x} = 2$.
At $x = 0$, $f(0) = 2$.
Now, check continuity at $x = 0$:
$\lim_{x \to 0^+} f(x) = 0$.
$\lim_{x \to 0^-} f(x) = 2$.
Since $\lim_{x \to 0^+} f(x) \neq \lim_{x \to 0^-} f(x)$, the limit does not exist at $x = 0$, so $f$ is discontinuous at $x = 0$.
For $x > 0$, $f(x) = 0$ (a constant function), which is continuous.
For $x < 0$, $f(x) = 2$ (a constant function), which is continuous.
Thus, $f$ is continuous for all $x$ except at $x = 0$.

(C) For $f(x)$ to be continuous at $x = 0$, we must have $k = \lim_{x \to 0} f(x)$.
$\lim_{x \to 0} \frac{(8-2x)^{1/3} - 2}{3 - (243+5x)^{1/5}} = \lim_{x \to 0} \frac{2(1 - x/4)^{1/3} - 2}{3 - 3(1 + 5x/243)^{1/5}}$
$= \lim_{x \to 0} \frac{2[1 + \frac{1}{3}(-x/4) + \dots] - 2}{3 - 3[1 + \frac{1}{5}(5x/243) + \dots]}$
$= \lim_{x \to 0} \frac{2 - x/6 - 2}{3 - 3 - x/81} = \lim_{x \to 0} \frac{-x/6}{-x/81} = \frac{81}{6} = \frac{27}{2}$.

(D) Since $f(x)$ is continuous at $x = 0$, we have $\lim_{x \to 0} f(x) = f(0)$.
$\lim_{x \to 0} \frac{8^x - 2^x}{k^x - 1} = 2$.
Divide numerator and denominator by $x$: $\lim_{x \to 0} \frac{\frac{8^x - 1}{x} - \frac{2^x - 1}{x}}{\frac{k^x - 1}{x}} = 2$.
Using the standard limit $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$, we get $\frac{\ln 8 - \ln 2}{\ln k} = 2$.
$\frac{\ln(8/2)}{\ln k} = 2 \implies \frac{\ln 4}{\ln k} = 2$.
$\ln 4 = 2 \ln k \implies \ln 4 = \ln(k^2)$.
$k^2 = 4 \implies k = 2$ (since $k^x$ is defined for $k > 0$ and $k \neq 1$ for the limit to exist).

(C) Since $f$ is continuous at $x = 1$, $f(1) = \lim_{x \to 1} f(x)$.
Let $x = 1 + h$. As $x \to 1$, $h \to 0$.
$f(1) = \lim_{h \to 0} \frac{1 + \cos(\pi(1+h))}{\pi(1-(1+h))^2}$.
Using $\cos(\pi + \theta) = -\cos(\theta)$, we get $\cos(\pi + \pi h) = -\cos(\pi h)$.
$f(1) = \lim_{h \to 0} \frac{1 - \cos(\pi h)}{\pi h^2}$.
Using the identity $1 - \cos(\theta) = 2\sin^2(\theta/2)$, we get $1 - \cos(\pi h) = 2\sin^2(\frac{\pi h}{2})$.
$f(1) = \lim_{h \to 0} \frac{2\sin^2(\frac{\pi h}{2})}{\pi h^2} = \lim_{h \to 0} \frac{2}{\pi} \left( \frac{\sin(\frac{\pi h}{2})}{\frac{\pi h}{2} \cdot \frac{2}{\pi}} \right)^2 = \lim_{h \to 0} \frac{2}{\pi} \cdot \frac{\pi^2}{4} \cdot \left( \frac{\sin(\frac{\pi h}{2})}{\frac{\pi h}{2}} \right)^2$.
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we get $f(1) = \frac{2}{\pi} \cdot \frac{\pi^2}{4} \cdot 1^2 = \frac{\pi}{2}$.

(A) Step $1$: Find the derivative $f'(x)$.
$f'(x) = \begin{cases} 2ax & \text{if } x < -1 \\ 2bx + a & \text{if } x > -1 \end{cases}$
Step $2$: Since $f'(x)$ is continuous everywhere, it must be continuous at $x = -1$. Thus, $\lim_{x \to -1^-} f'(x) = \lim_{x \to -1^+} f'(x)$.
$2a(-1) = 2b(-1) + a \implies -2a = -2b + a \implies 3a = 2b \implies b = \frac{3}{2}a$.
Step $3$: For $f'(x)$ to exist at $x = -1$, $f(x)$ must be continuous at $x = -1$. Thus, $\lim_{x \to -1^-} f(x) = f(-1)$.
$a(-1)^2 + b = b(-1)^2 + a(-1) + 4 \implies a + b = b - a + 4 \implies 2a = 4 \implies a = 2$.
Step $4$: Substitute $a = 2$ into $b = \frac{3}{2}a$.
$b = \frac{3}{2}(2) = 3$. Therefore, $a = 2, b = 3$.

(C) Since the function $f$ is continuous at $x = \pi$, we have $f(\pi) = \lim_{x \to \pi} f(x)$.
Let $t = x - \pi$. As $x \to \pi$, $t \to 0$.
Then $f(\pi) = \lim_{t \to 0} \frac{1 - \cos(7t)}{5t^2}$.
Using the identity $1 - \cos(\theta) = 2 \sin^2(\theta/2)$, we get $1 - \cos(7t) = 2 \sin^2(7t/2)$.
$f(\pi) = \lim_{t \to 0} \frac{2 \sin^2(7t/2)}{5t^2}$.
$f(\pi) = \frac{2}{5} \lim_{t \to 0} \left( \frac{\sin(7t/2)}{t} \right)^2$.
Multiply and divide by $(7/2)^2$: $f(\pi) = \frac{2}{5} \times (7/2)^2 \times \lim_{t \to 0} \left( \frac{\sin(7t/2)}{7t/2} \right)^2$.
Since $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we have $f(\pi) = \frac{2}{5} \times \frac{49}{4} \times 1^2 = \frac{49}{10}$.

(C) function $f(x)$ is continuous at $x=0$ if $\lim_{x \to 0} f(x) = f(0)$.
$(A)$ $\lim_{x \to 0} (1+x)^{\frac{2}{x}} = e^{\lim_{x \to 0} \frac{2}{x} \cdot x} = e^2 = f(0)$. Continuous.
$(B)$ $\lim_{x \to 0} (\sin x - \cos x) = 0 - 1 = -1 = f(0)$. Continuous.
$(C)$ For $x \to 0^+$, $\lim_{x \to 0^+} \frac{e^{1/x}(1 - e^{-1/x})}{e^{1/x}(1 + e^{-1/x})} = 1$. For $x \to 0^-$, $\lim_{x \to 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} = \frac{0 - 1}{0 + 1} = -1$. Since $\text{LHL} \neq \text{RHL}$, the limit does not exist. Discontinuous.
$(D)$ $\lim_{x \to 0} \frac{e^{5x} - e^{2x}}{\sin 3x} = \lim_{x \to 0} \frac{\frac{e^{5x}-1}{x} - \frac{e^{2x}-1}{x}}{\frac{\sin 3x}{3x} \cdot 3} = \frac{5 - 2}{3} = 1 = f(0)$. Continuous.

(A) Step $1$: Simplify the numerator using $\sin A + \sin B = 2 \sin \frac{A+B}{2} \cos \frac{A-B}{2}$.
$\sin 3x + \sin x = 2 \sin 2x \cos x$.
Numerator $= 4\sqrt{2} (2 \sin 2x \cos x) = 8\sqrt{2} \sin 2x \cos x$.
Step $2$: Simplify the denominator using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$ and $\cos C - \cos D = -2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$.
$2 \sin 2x \sin \frac{3x}{2} = \cos(2x - \frac{3x}{2}) - \cos(2x + \frac{3x}{2}) = \cos \frac{x}{2} - \cos \frac{7x}{2}$.
Denominator $= \cos \frac{x}{2} - \cos \frac{7x}{2} + \cos \frac{5x}{2} - \cos \frac{3x}{2} = (\cos \frac{x}{2} - \cos \frac{3x}{2}) + (\cos \frac{5x}{2} - \cos \frac{7x}{2})$.
$= 2 \sin x \sin \frac{x}{2} + 2 \sin 3x \sin \frac{x}{2} = 2 \sin \frac{x}{2} (\sin x + \sin 3x) = 2 \sin \frac{x}{2} (2 \sin 2x \cos x) = 4 \sin \frac{x}{2} \sin 2x \cos x$.
Step $3$: Calculate $f(x) = \frac{8\sqrt{2} \sin 2x \cos x}{4 \sin \frac{x}{2} \sin 2x \cos x} = \frac{2\sqrt{2}}{\sin \frac{x}{2}}$.
Step $4$: For continuity at $x = \frac{\pi}{2}$, $f(\frac{\pi}{2}) = \lim_{x \to \frac{\pi}{2}} \frac{2\sqrt{2}}{\sin \frac{x}{2}} = \frac{2\sqrt{2}}{\sin \frac{\pi}{4}} = \frac{2\sqrt{2}}{1/\sqrt{2}} = 4$.

(D) rational function is discontinuous where its denominator is zero.
Set the denominator to zero: $x^2 - 5|x| + 6 = 0$.
Since $x^2 = |x|^2$, we can write this as $|x|^2 - 5|x| + 6 = 0$.
Let $|x| = t$, then $t^2 - 5t + 6 = 0$.
Factoring the quadratic: $(t - 2)(t - 3) = 0$.
So, $|x| = 2$ or $|x| = 3$.
This gives $x = \pm 2$ or $x = \pm 3$.
The points of discontinuity are $x = 2, -2, 3, -3$.
There are $4$ such points.

(A) For $f(x)$ to be continuous at $x = 0$, $f(0) = \lim_{x \to 0} f(x)$.
Given $f(x) = \frac{3^{x+3} - 3^{-x} - 2}{\tan x \cdot \log(1+x)}$.
Rewrite the numerator: $3^{x+3} - 3^{-x} - 2 = 27 \cdot 3^x - 3^{-x} - 2$.
As $x \to 0$, the expression is of the form $\frac{27-1-2}{0} = \frac{24}{0}$, which is undefined. Re-evaluating the expression: $3^{x+3} - 3^{-x} - 2 = 3^x \cdot 27 - \frac{1}{3^x} - 2 = \frac{27 \cdot 3^{2x} - 2 \cdot 3^x - 1}{3^x}$.
Let $t = 3^x$. As $x \to 0, t \to 1$. The numerator becomes $27t^2 - 2t - 1 = (9t+1)(3t-1)$.
Thus, $\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(9 \cdot 3^x + 1)(3 \cdot 3^x - 1)}{3^x \cdot \tan x \cdot \log(1+x)}$.
Using $\lim_{x \to 0} \frac{\tan x}{x} = 1$ and $\lim_{x \to 0} \frac{\log(1+x)}{x} = 1$, we divide numerator and denominator by $x^2$:
$f(0) = \lim_{x \to 0} \frac{(9 \cdot 3^x + 1) \cdot \frac{3 \cdot 3^x - 1}{x}}{3^x \cdot \frac{\tan x}{x} \cdot \frac{\log(1+x)}{x} \cdot x}$.
This limit tends to infinity. Note: If the numerator was $3^{x+1} - 3^{-x} - 2$, then $3(3^x) - 1/3^x - 2 = (3 \cdot 3^{2x} - 2 \cdot 3^x - 1)/3^x = (3 \cdot 3^x + 1)(3^x - 1)/3^x$. Then $\lim_{x \to 0} \frac{(3 \cdot 3^x + 1)(3^x - 1)}{3^x \cdot x^2} = \frac{4 \cdot \ln 3}{1} = 4 \ln 3$. Given the options, the intended numerator is $3^{x+1} - 3^{-x} - 2$ and the denominator $\tan x \cdot x$. The result is $4 \ln 3$.

(B) For $f(x)$ to be continuous at $x = 2$, $f(2) = \lim_{x \to 2} f(x)$.
Let $L = \lim_{x \to 2} \left( \frac{5x - 8}{8 - 3x} \right)^{\frac{3}{2x-4}}$.
As $x \to 2$, the base $\frac{5(2)-8}{8-3(2)} = \frac{2}{2} = 1$ and the exponent $\frac{3}{2(2)-4} \to \infty$.
This is a $1^\infty$ form. We use the formula $\lim_{x \to a} [g(x)]^{h(x)} = e^{\lim_{x \to a} h(x)[g(x)-1]}$.
$L = e^{\lim_{x \to 2} \frac{3}{2x-4} \left( \frac{5x-8}{8-3x} - 1 \right)}$.
Simplify the term inside the bracket: $\frac{5x-8 - (8-3x)}{8-3x} = \frac{8x-16}{8-3x} = \frac{8(x-2)}{8-3x}$.
Substitute back: $L = e^{\lim_{x \to 2} \frac{3}{2(x-2)} \cdot \frac{8(x-2)}{8-3x}}$.
Cancel $(x-2)$: $L = e^{\lim_{x \to 2} \frac{24}{2(8-3x)}}$.
Evaluate the limit: $L = e^{\frac{24}{2(8-6)}} = e^{\frac{24}{4}} = e^6$.

(B) Step $1$: Evaluate the limit of $f(x)$ as $x \to 0$.
Step $2$: For $x$ very close to $0$, $\tan x$ is very close to $0$, so $\tan^2 x$ is a small positive value close to $0$.
Step $3$: Since $0 < \tan^2 x < 1$ for $x \in (-\epsilon, \epsilon) \setminus \{0\}$, the greatest integer function $[\tan^2 x] = 0$.
Step $4$: Thus, $\lim_{x \to 0} f(x) = 0$.
Step $5$: Since $f(0) = [\tan^2 0] = [0] = 0$, we have $\lim_{x \to 0} f(x) = f(0) = 0$, so $f(x)$ is continuous at $x = 0$.
Step $6$: Since $f(x) = 0$ in a neighborhood of $x = 0$, $f'(0) = 0$.

(A) For $f(x)$ to be continuous at $x = 3$, the left-hand limit must equal the right-hand limit and the value of the function at $x = 3$.
$\lim_{x \to 3^-} f(x) = f(3) = 3a + 1$.
$\lim_{x \to 3^+} f(x) = 3b + 3$.
Equating the limits: $3a + 1 = 3b + 3$.
$3a - 3b = 3 - 1$.
$3(a - b) = 2$.
$a - b = \frac{2}{3}$.

(B) Step $1$: Calculate $\lim_{x \to 1} f(x)$. Since $1 < 2$, we use $f(x) = x + 1$. Thus, $\lim_{x \to 1} (x + 1) = 1 + 1 = 2$.
Step $2$: Calculate $\lim_{x \to 2} f(x)$. We check the left-hand and right-hand limits.
Left-hand limit: $\lim_{x \to 2^-} f(x) = \lim_{x \to 2} (x + 1) = 2 + 1 = 3$.
Right-hand limit: $\lim_{x \to 2^+} f(x) = \lim_{x \to 2} (x^2 - 1) = 2^2 - 1 = 4 - 1 = 3$.
Since both limits are equal, $\lim_{x \to 2} f(x) = 3$.
Step $3$: The sum is $\lim_{x \to 1} f(x) + \lim_{x \to 2} f(x) = 2 + 3 = 5$.

(D) For $f(x)$ to be continuous at $x = 1$, the condition $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$ must hold.
Step $1$: Calculate $f(1) = 3(1) - 1 = 2$.
Step $2$: Calculate the left-hand limit: $\lim_{x \to 1^-} (ax + 7) = a(1) + 7 = a + 7$. Equating to $f(1)$, we get $a + 7 = 2$, which implies $a = -5$.
Step $3$: Calculate the right-hand limit: $\lim_{x \to 1^+} \frac{x + 3}{b} = \frac{1 + 3}{b} = \frac{4}{b}$. Equating to $f(1)$, we get $\frac{4}{b} = 2$, which implies $b = 2$.
Thus, $a = -5$ and $b = 2$.