The problem consists of four parts: a) Compute two limits: (i) $\lim_{x \to 0} \frac{\sin^{2025}(-x)}{x^{2025}}$ (ii) $\lim_{x \to -\infty} \frac{\sqrt{x^2 + 2025}}{2x + 2025}$ b) Use logarithmic differentiation to find the derivative $y'$ of the function $y = 3^{x^3} \sqrt{x} (x^4 + 1)^5$. c) Determine if there exists a real number that is exactly one less than its square root. If it exists, find it. If not, explain why. d) A cup of coffee cools from 98°C to 65°C in a room at 25°C, while cooling at 4°C per minute. Determine when the coffee's temperature is decreasing at this rate, giving the answer to the nearest minute.

1. Problem Description

The problem consists of four parts:

a) Compute two limits:

(i) $\lim_{x \to 0} \frac{\sin^{2025}(-x)}{x^{2025}}$

(ii) $\lim_{x \to -\infty} \frac{\sqrt{x^2 + 2025}}{2x + 2025}$

b) Use logarithmic differentiation to find the derivative $y'$ of the function $y = 3^{x^3} \sqrt{x} (x^4 + 1)^5$.

c) Determine if there exists a real number that is exactly one less than its square root. If it exists, find it. If not, explain why.

d) A cup of coffee cools from 98°C to 65°C in a room at 25°C, while cooling at 4°C per minute. Determine when the coffee's temperature is decreasing at this rate, giving the answer to the nearest minute.

2. Solution Steps

a) (i)

We can rewrite the limit as:

$\lim_{x \to 0} \frac{\sin^{2025}(-x)}{x^{2025}} = \lim_{x \to 0} \left(\frac{\sin(-x)}{x}\right)^{2025} = \lim_{x \to 0} \left(-\frac{\sin(x)}{x}\right)^{2025}$

Since $\lim_{x \to 0} \frac{\sin(x)}{x} = 1$, we have

$\lim_{x \to 0} \left(-\frac{\sin(x)}{x}\right)^{2025} = (-1)^{2025} = -1$

a) (ii)

$\lim_{x \to -\infty} \frac{\sqrt{x^2 + 2025}}{2x + 2025}$

As $x \to -\infty$, we can approximate $\sqrt{x^2 + 2025} \approx \sqrt{x^2} = |x| = -x$ (since $x < 0$).

Thus, we have

$\lim_{x \to -\infty} \frac{-x}{2x + 2025} = \lim_{x \to -\infty} \frac{-x}{x(2 + \frac{2025}{x})} = \lim_{x \to -\infty} \frac{-1}{2 + \frac{2025}{x}}$

As $x \to -\infty$, $\frac{2025}{x} \to 0$. Thus, the limit is $\frac{-1}{2 + 0} = -\frac{1}{2}$

b) $y = 3^{x^3} \sqrt{x} (x^4 + 1)^5 = 3^{x^3} x^{1/2} (x^4 + 1)^5$

Take the natural logarithm of both sides:

$\ln y = \ln \left(3^{x^3} x^{1/2} (x^4 + 1)^5\right) = \ln(3^{x^3}) + \ln(x^{1/2}) + \ln((x^4 + 1)^5)$

$\ln y = x^3 \ln 3 + \frac{1}{2} \ln x + 5 \ln(x^4 + 1)$

Differentiate both sides with respect to $x$:

$\frac{1}{y} \frac{dy}{dx} = 3x^2 \ln 3 + \frac{1}{2} \cdot \frac{1}{x} + 5 \cdot \frac{4x^3}{x^4 + 1} = 3x^2 \ln 3 + \frac{1}{2x} + \frac{20x^3}{x^4 + 1}$

$\frac{dy}{dx} = y \left(3x^2 \ln 3 + \frac{1}{2x} + \frac{20x^3}{x^4 + 1}\right)$

$y' = 3^{x^3} \sqrt{x} (x^4 + 1)^5 \left(3x^2 \ln 3 + \frac{1}{2x} + \frac{20x^3}{x^4 + 1}\right)$

c) Let $x$ be the real number. Then $x = \sqrt{x} - 1$.

$\sqrt{x} = x + 1$

Square both sides:

$x = (x + 1)^2 = x^2 + 2x + 1$

$x^2 + x + 1 = 0$

The discriminant is $b^2 - 4ac = 1^2 - 4(1)(1) = 1 - 4 = -3 < 0$.

Since the discriminant is negative, there are no real solutions.

d) Newton's Law of Cooling:

$\frac{dT}{dt} = k(T - T_s)$, where $T$ is the temperature of the coffee, $T_s$ is the surrounding temperature (25°C), and $k$ is a constant.

We are given that when $T = 65°C$, $\frac{dT}{dt} = -4°C$ per minute.

$-4 = k(65 - 25) = 40k$

$k = -\frac{4}{40} = -\frac{1}{10}$

The question is, when is the temperature decreasing at 4°C per minute? We already know it is at 65°C.

Thus the question doesn't make sense and no calculation is needed. But we interpret as at what time is the coffee at 65 degrees. However, to calculate the time to reach 65, more information would be needed. We cannot determine the time when the coffee is at 65 degrees based on given information.

3. Final Answer

a) (i) -1

a) (ii) -1/2

b) $y' = 3^{x^3} \sqrt{x} (x^4 + 1)^5 \left(3x^2 \ln 3 + \frac{1}{2x} + \frac{20x^3}{x^4 + 1}\right)$

c) No such real number exists, because the quadratic equation $x^2 + x + 1 = 0$ has no real roots.

d) At 65°C. (We cannot determine the time)