Limits and Continuity

• Step 1: Evaluate Left-Hand Limit (LHL)
  Let $x = 0 - h$, where $h \to 0^+$. Since $x < 0$, we have $|x| = -x$. \[ \text{LHL} = \lim_{h \to 0^+} \frac{-h}{|-h|} = \lim_{h \to 0^+} \frac{-h}{h} = -1 \]
• Step 2: Evaluate Right-Hand Limit (RHL)
  Let $x = 0 + h$, where $h \to 0^+$. Since $x > 0$, we have $|x| = x$. \[ \text{RHL} = \lim_{h \to 0^+} \frac{h}{|h|} = \lim_{h \to 0^+} \frac{h}{h} = 1 \]

• Step 1: Evaluate LHL
  Let $x = 1 - h$, where $h$ is an vanishingly small positive number. This places $x$ in the interval $0 \le x < 1$. By definition, the floor value of any number in this interval is 0: \[ \text{LHL} = \lim_{h \to 0^+} [1 - h] = 0 \]
• Step 2: Evaluate RHL
  Let $x = 1 + h$, where $h \to 0^+$. This places $x$ in the interval $1 \le x < 2$. By definition, the floor value of any number in this interval is 1: \[ \text{RHL} = \lim_{h \to 0^+} [1 + h] = 1 \]

• Step 1: Set up the LHL equation
  \[ \text{LHL} = \lim_{x \to 1^-} (ax + b) = a(1) + b = a + b \] Since the limit must equal $f(1)$, set: $a + b = 4 \quad \text{--- (Equation 1)}$
• Step 2: Set up the RHL equation
  \[ \text{RHL} = \lim_{x \to 1^+} (2ax^2 - b) = 2a(1)^2 - b = 2a - b \] Set this equal to $f(1)$: $2a - b = 4 \quad \text{--- (Equation 2)}$

Direct substitution of $x = 0$ yields $\frac{\ln(1)}{0} = \frac{0}{0}$, which is an indeterminate form. We can resolve this by using algebraic manipulation to introduce standard limit structures: \[ \lim_{x \to 0} \frac{\ln(1 + 3x)}{\sin(5x)} = \lim_{x \to 0} \left[ \frac{\ln(1 + 3x)}{3x} \cdot \frac{5x}{\sin(5x)} \cdot \frac{3x}{5x} \right] \] By applying the limit distribution properties, we can evaluate each fraction independently: \[ \lim_{x \to 0} \frac{\ln(1 + 3x)}{3x} = 1 \quad (\text{since } 3x \to 0) \] \[ \lim_{x \to 0} \frac{5x}{\sin(5x)} = \frac{1}{\lim_{5x \to 0} \frac{\sin(5x)}{5x}} = \frac{1}{1} = 1 \] Multiply the simplified terms by the remaining constant scaling factor: \[ \lim_{x \to 0} \left[ 1 \cdot 1 \cdot \frac{3}{5} \right] = \frac{3}{5} \] Final Answer: $\frac{3}{5}$.

As $x \to 0$, the base $\frac{\sin x}{x} \to 1$ and the exponent $\frac{1}{x^2} \to \infty$. This matches the standard $1^\infty$ indeterminate form. Apply the shortcut transformation theorem: $\lim_{x \to a} [f(x)]^{g(x)} = e^{\lim_{x \to a} (f(x) - 1)g(x)}$. \[ L = e^{\lim_{x \to 0} \left( \frac{\sin x}{x} - 1 \right) \cdot \frac{1}{x^2}} = e^{\lim_{x \to 0} \left( \frac{\sin x - x}{x^3} \right)} \] To evaluate the limit in the exponent, substitute the Taylor series expansion for the sine function ($\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \dots$): \[ \lim_{x \to 0} \frac{\left(x - \frac{x^3}{6} + \mathcal{O}(x^5)\right) - x}{x^3} = \lim_{x \to 0} \frac{-\frac{x^3}{6} + \mathcal{O}(x^5)}{x^3} = -\frac{1}{6} \] Substitute this calculated value back into the exponent: \[ L = e^{-1/6} = \frac{1}{\sqrt[6]{e}} \] Final Answer: $e^{-1/6}$.

• Numerator: $n = \frac{2}{3} \implies \frac{2}{3} a^{2/3 - 1} = \frac{2}{3} a^{-1/3}$
• Denominator: $n = \frac{3}{4} \implies \frac{3}{4} a^{3/4 - 1} = \frac{3}{4} a^{-1/4}$

• Numerator Polynomial: $x^3 - 3x^2 + 4 = (x-2)(x^2 - x - 2) = (x-2)(x-2)(x+1) = (x-2)^2(x+1)$
• Denominator Polynomial: $x^4 - 4x^3 + 16x - 16$. Notice that we can group terms: \[ (x^4 - 16) - 4x(x^2 - 4) = (x^2 - 4)(x^2 + 4) - 4x(x^2 - 4) = (x^2 - 4)(x^2 - 4x + 4) \] Substitute the factored parts: \[ (x-2)(x+2)(x-2)^2 = (x-2)^3(x+2) \]

Direct substitution yields $\frac{0}{0}$. Multiply both the numerator and the denominator by the radical conjugate expression $\sqrt{1 + x + x^2} + 1$: \[ \lim_{x \to 0} \frac{\left(\sqrt{1 + x + x^2} - 1\right)\left(\sqrt{1 + x + x^2} + 1\right)}{x \left(\sqrt{1 + x + x^2} + 1\right)} \] Simplify the numerator using the difference of squares identity ($(a-b)(a+b) = a^2 - b^2$): \[ \lim_{x \to 0} \frac{(1 + x + x^2) - 1}{x \left(\sqrt{1 + x + x^2} + 1\right)} = \lim_{x \to 0} \frac{x + x^2}{x \left(\sqrt{1 + x + x^2} + 1\right)} \] Factor out $x$ from the numerator to cancel it with the $x$ in the denominator: \[ \lim_{x \to 0} \frac{x(1 + x)}{x \left(\sqrt{1 + x + x^2} + 1\right)} = \lim_{x \to 0} \frac{1 + x}{\sqrt{1 + x + x^2} + 1} \] Now evaluate the limit by direct substitution: \[ \frac{1 + 0}{\sqrt{1 + 0 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2} \] Final Answer: $\frac{1}{2}$.

Direct substitution yields $\frac{1 - 1 - 0}{0 - 0} = \frac{0}{0}$. Since both the numerator and denominator are differentiable near 0, we can apply L'Hôpital's Rule by differentiating the top and bottom independently: \[ \text{Numerator derivative: } \frac{d}{dx}(e^x - e^{-x} - 2x) = e^x + e^{-x} - 2 \] \[ \text{Denominator derivative: } \frac{d}{dx}(x - \sin x) = 1 - \cos x \] This gives the new limit expression: \[ \lim_{x \to 0} \frac{e^x + e^{-x} - 2}{1 - \cos x} \] Substitute $x = 0$: $\frac{1 + 1 - 2}{1 - 1} = \frac{0}{0}$. Since the expression is still indeterminate, apply L'Hôpital's Rule a second time: \[ \lim_{x \to 0} \frac{\frac{d}{dx}(e^x + e^{-x} - 2)}{\frac{d}{dx}(1 - \cos x)} = \lim_{x \to 0} \frac{e^x - e^{-x}}{\sin x} \] Substitute $x = 0$: $\frac{0}{0}$. Apply L'Hôpital's Rule a third time: \[ \lim_{x \to 0} \frac{\frac{d}{dx}(e^x - e^{-x})}{\frac{d}{dx}(\sin x)} = \lim_{x \to 0} \frac{e^x + e^{-x}}{\cos x} \] Substitute $x = 0$ into this final expression: \[ \frac{e^0 + e^0}{\cos 0} = \frac{1 + 1}{1} = 2 \] Final Answer: $2$.

The sine function is bounded between $-1$ and $1$ for all real inputs ($-\times \le \sin(x) \le 1$). Let us use this inequality to build upper and lower bounding functions for the complete expression. Multiply the inequality by the positive value $x^2$: \[ -x^2 \le x^2 \sin(x) \le x^2 \] Add $2x$ across all parts of the inequality: \[ -x^2 + 2x \le x^2 \sin(x) + 2x \le x^2 + 2x \] Divide all parts by the positive denominator $x^3 + 1$: \[ \frac{-x^2 + 2x}{x^3 + 1} \le \frac{x^2 \sin(x) + 2x}{x^3 + 1} \le \frac{x^2 + 2x}{x^3 + 1} \] Now evaluate the limit as $x \to \infty$ for the lower and upper bounding functions using dominant term analysis: \[ \lim_{x \to \infty} \frac{-x^2 + 2x}{x^3 + 1} = \lim_{x \to \infty} \frac{-1/x + 2/x^2}{1 + 1/x^3} = \frac{0}{1} = 0 \] \[ \lim_{x \to \infty} \frac{x^2 + 2x}{x^3 + 1} = \lim_{x \to \infty} \frac{1/x + 2/x^2}{1 + 1/x^3} = \frac{0}{1} = 0 \] Since both the upper and lower bounding functions converge to 0, the Sandwich Theorem guarantees that the central function must converge to the same value. Final Answer: $0$.

• $x \cos x = x \left( 1 - \frac{x^2}{2} + \mathcal{O}(x^4) \right) = x - \frac{x^3}{2} + \mathcal{O}(x^5)$
• $\sin x = x - \frac{x^3}{6} + \mathcal{O}(x^5)$
• $x^2 \ln(1+x) = x^2 \left( x - \frac{x^2}{2} + \mathcal{O}(x^3) \right) = x^3 - \frac{x^4}{2} + \mathcal{O}(x^5)$

We can frame the floor function using the standard bounding inequality: $z - 1 < [z] \le z$. Apply this inequality to each term in the numerator: \[ (1x - 1) < [1x] \le 1x \] \[ (2x - 1) < [2x] \le 2x \] \[ \dots \] \[ (nx - 1) < [nx] \le nx \] Sum all $n$ inequalities together: \[ \sum_{k=1}^n (kx - 1) < \sum_{k=1}^n [kx] \le \sum_{k=1}^n kx \] Factor out the constants to evaluate the arithmetic sums: \[ x \cdot \frac{n(n+1)}{2} - n < \sum_{k=1}^n [kx] \le x \cdot \frac{n(n+1)}{2} \] Divide the entire inequality by $n^2$: \[ \frac{x n(n+1)}{2n^2} - \frac{n}{n^2} < \frac{\sum_{k=1}^n [kx]}{n^2} \le \frac{x n(n+1)}{2n^2} \] Evaluate the limit as $n \to \infty$ for the lower and upper bounding functions independently: \[ \text{Upper Limit: } \lim_{n \to \infty} \frac{x(n^2 + n)}{2n^2} = \frac{x}{2} \] \[ \text{Lower Limit: } \lim_{n \to \infty} \left( \frac{x(n^2+n)}{2n^2} - \frac{1}{n} \right) = \frac{x}{2} - 0 = \frac{x}{2} \] Since the upper and lower bounding functions converge to the exact same value, the Sandwich Theorem guarantees that the central series limit must converge to $\frac{x}{2}$. Final Answer: $\frac{x}{2}$.

For $f(x)$ to be continuous at $x = 0$, the limit as $x \to 0$ must equal the functional assignment value: $\lim_{x \to 0} f(x) = f(0) = k$. Let us evaluate the limit: \[ \lim_{x \to 0} \frac{1 - \cos(4x)}{8x^2} \] Substitute the double-angle identity $1 - \cos(2\theta) = 2\sin^2\theta$, where $2\theta = 4x \implies \theta = 2x$: \[ 1 - \cos(4x) = 2\sin^2(2x) \] Substitute this back into the limit expression: \[ \lim_{x \to 0} \frac{2\sin^2(2x)}{8x^2} = \lim_{x \to 0} \frac{\sin^2(2x)}{4x^2} \] Group the square powers together to form a standard trigonometric limit structure: \[ \lim_{x \to 0} \left( \frac{\sin(2x)}{2x} \right)^2 \] Since $x \to 0 \implies 2x \to 0$, apply the standard limit identity $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$: \[ \lim_{x \to 0} \left( \frac{\sin(2x)}{2x} \right)^2 = (1)^2 = 1 \] Therefore, for the function to be continuous, set $k = 1$. Final Answer: $k = 1$.

• Step 1: Calculate $f(1)$ and RHL
  The equality rule for $x = 1$ belongs to the upper branch: $f(1) = |1 - 3| = |-2| = 2$. The right-hand limit approaches from values greater than 1, so it uses the same branch: \[ \text{RHL} = \lim_{x \to 1^+} |x - 3| = |1 - 3| = 2 \]
• Step 2: Calculate LHL
  The left-hand limit approaches from values less than 1, so it uses the lower branch: \[ \text{LHL} = \lim_{x \to 1^-} \left( \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4} \right) = \frac{1^2}{4} - \frac{3(1)}{2} + \frac{13}{4} \] Combine the fractions using a common denominator: \[ \text{LHL} = \frac{1}{4} - \frac{6}{4} + \frac{13}{4} = \frac{1 - 6 + 13}{4} = \frac{8}{4} = 2 \]

• Step 1: Calculate $f(2)$
  Substitute $x = 2$ directly, noting that for integers, $[2] = 2$ and $[-2] = -2$: \[ f(2) = 2 - 2 - (-2) = 0 + 2 = 2 \]
• Step 2: Calculate LHL
  Let $x = 2 - h$, where $h \to 0^+$. This places $x$ in the interval $1 < x < 2 \implies [x] = 1$. The negative term becomes $-x = -2 + h$, which places it in the interval $-2 < -x < -1 \implies [-x] = -2$. \[ \text{LHL} = \lim_{h \to 0^+} \left( [2-h] - (2-h) - [-(2-h)] \right) = 1 - 2 - (-2) = -1 + 2 = 1 \]

Let us check the function against the continuity criteria at $x = 3$. First, evaluate the function at the point: $f(3) = \frac{3^2 - 9}{3 - 3} = \frac{0}{0}$, which is undefined. Now evaluate the two-sided limit as $x \to 3$ by factoring the numerator polynomial: \[ \lim_{x Caps 3} \frac{x^2 - 9}{x - 3} = \lim_{x \to 3} \frac{(x-3)(x+3)}{x - 3} \] Since $x \to 3 \implies x \neq 3$, we can cancel the common factor $(x-3)$: \[ \lim_{x \to 3} (x + 3) = 3 + 3 = 6 \] The two-sided limit exists and is a finite real number ($L = 6$), but the function itself is undefined at $x = 3$. This fits the definition of a removable discontinuity. We can remove it and make the function continuous by defining $f(3) = 6$. Final Answer: Removable Discontinuity.

• Step 1: Calculate Left-Hand Limit (LHL)
  Let $x \to 0^- \implies \frac{1}{x} \to -\infty$. Therefore, $e^{1/x} \to e^{-\infty} = 0$. Substitute this limiting value into the expression: \[ \text{LHL} = \frac{0 - 1}{0 + 1} = -1 \]
• Step 2: Calculate Right-Hand Limit (RHL)
  Let $x \to 0^+ \implies \frac{1}{x} \to \infty$. Therefore, $e^{1/x} \to \infty$. To evaluate this $\frac{\infty}{\infty}$ indeterminate form, divide both the numerator and denominator by $e^{1/x}$: \[ \text{RHL} = \lim_{x \to 0^+} \frac{1 - e^{-1/x}}{1 + e^{-1/x}} \] As $x \to 0^+ \implies -1/x \to -\infty \implies e^{-1/x} \to 0$. \[ \text{RHL} = \frac{1 - 0}{1 + 0} = 1 \]

Let us define the polynomial function $f(x) = x^5 - 3x - 1$. Since $f(x)$ is a polynomial, it is continuous everywhere, including on the closed interval $[1, 2]$. Now evaluate the value of the function at the interval endpoints $x = 1$ and $x = 2$: \[ f(1) = 1^5 - 3(1) - 1 = 1 - 3 - 1 = -3 \quad (\text{which is } < 0) \] \[ f(2) = 2^5 - 3(2) - 1 = 32 - 6 - 1 = 25 \quad (\text{which is } > 0) \] Calculate the product of the endpoint values: \[ f(1) \cdot f(2) = (-3) \times 25 = -75 < 0 \] The function values at the endpoints have opposite signs ($f(1) < 0$ and $f(2) > 0$). Since the function is continuous, the Intermediate Value Theorem (IVT) guarantees that the graph must cross the x-axis at least once between $x = 1$ and $x = 2$. Therefore, there exists at least one real number $c \in (1, 2)$ such that $f(c) = 0$. This confirms the existence of a real root. Final Answer: Proved.