We are asked to solve four problems: 2.1. Use the Intermediate Value Theorem to show that there is a root of the equation $\cos\sqrt{x} = e^x - 2$ in the interval $(0, 1)$. 2.2. Prove that $\lim_{x \to 10} (3 - \frac{4}{5}x) = -5$ using the precise definition of the limit. 2.3. Prove that $\lim_{x \to -1^-} \frac{5}{(x+1)^3} = -\infty$ using the precise definition of the limit. 2.4. Find the limit $\lim_{t \to 0} \frac{\tan^2 4t}{t^2}$.

1. Problem Description

We are asked to solve four problems:

2.

1. Use the Intermediate Value Theorem to show that there is a root of the equation $\cos\sqrt{x} = e^x - 2$ in the interval $(0, 1)$.

2.

2. Prove that $\lim_{x \to 10} (3 - \frac{4}{5}x) = -5$ using the precise definition of the limit.

2.

3. Prove that $\lim_{x \to -1^-} \frac{5}{(x+1)^3} = -\infty$ using the precise definition of the limit.

2.

4. Find the limit $\lim_{t \to 0} \frac{\tan^2 4t}{t^2}$.

2. Solution Steps

2.

1. Let $f(x) = \cos\sqrt{x} - e^x + 2$. We want to show that there exists $c \in (0, 1)$ such that $f(c) = 0$.

First, we evaluate $f(x)$ at the endpoints of the interval $(0, 1)$:

$f(0) = \cos\sqrt{0} - e^0 + 2 = \cos(0) - 1 + 2 = 1 - 1 + 2 = 2 > 0$.

$f(1) = \cos\sqrt{1} - e^1 + 2 = \cos(1) - e + 2 \approx 0.54 - 2.72 + 2 = -0.18 < 0$.

Since $f(0) > 0$ and $f(1) < 0$, and $f(x)$ is continuous on the interval $[0, 1]$, by the Intermediate Value Theorem, there exists a $c \in (0, 1)$ such that $f(c) = 0$. Thus, there is a root in the interval $(0, 1)$.

2.

2. We want to show that $\lim_{x \to 10} (3 - \frac{4}{5}x) = -5$.

Let $f(x) = 3 - \frac{4}{5}x$ and $L = -5$. We need to show that for every $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x - 10| < \delta$, then $|f(x) - L| < \epsilon$.

$|f(x) - L| = |3 - \frac{4}{5}x - (-5)| = |3 - \frac{4}{5}x + 5| = |8 - \frac{4}{5}x| = |\frac{40 - 4x}{5}| = \frac{4}{5}|10 - x| = \frac{4}{5}|x - 10|$.

We want $|f(x) - L| < \epsilon$, so $\frac{4}{5}|x - 10| < \epsilon$.

This implies $|x - 10| < \frac{5}{4}\epsilon$.

Thus, we can choose $\delta = \frac{5}{4}\epsilon$.

If $0 < |x - 10| < \delta = \frac{5}{4}\epsilon$, then $|f(x) - L| = \frac{4}{5}|x - 10| < \frac{4}{5}(\frac{5}{4}\epsilon) = \epsilon$.

Therefore, $\lim_{x \to 10} (3 - \frac{4}{5}x) = -5$.

2.

3. We want to prove that $\lim_{x \to -1^-} \frac{5}{(x+1)^3} = -\infty$.

Let $M > 0$ be given. We want to find a $\delta > 0$ such that if $-1 - \delta < x < -1$, then $\frac{5}{(x+1)^3} < -M$.

$\frac{5}{(x+1)^3} < -M$ implies $(x+1)^3 < -\frac{5}{M}$.

Taking the cube root of both sides, we have $x+1 < \sqrt[3]{-\frac{5}{M}} = -\sqrt[3]{\frac{5}{M}}$.

Thus, $x < -1 - \sqrt[3]{\frac{5}{M}}$.

We want $-1 - \delta < x < -1$, so we can choose $\delta = \sqrt[3]{\frac{5}{M}}$.

If $-1 - \delta < x < -1$, then $-1 - \sqrt[3]{\frac{5}{M}} < x < -1$, which implies $x+1 < -\sqrt[3]{\frac{5}{M}}$.

Then $(x+1)^3 < -\frac{5}{M}$, so $\frac{5}{(x+1)^3} < -M$.

Therefore, $\lim_{x \to -1^-} \frac{5}{(x+1)^3} = -\infty$.

2.

4. We want to find the limit $\lim_{t \to 0} \frac{\tan^2 4t}{t^2}$.

$\lim_{t \to 0} \frac{\tan^2 4t}{t^2} = \lim_{t \to 0} \frac{\tan 4t}{t} \cdot \frac{\tan 4t}{t} = \lim_{t \to 0} \frac{\tan 4t}{4t} \cdot 4 \cdot \frac{\tan 4t}{4t} \cdot 4 = \lim_{t \to 0} (\frac{\tan 4t}{4t})^2 \cdot 16$.

Since $\lim_{x \to 0} \frac{\tan x}{x} = 1$, we have $\lim_{t \to 0} \frac{\tan 4t}{4t} = 1$.

Therefore, $\lim_{t \to 0} \frac{\tan^2 4t}{t^2} = 1^2 \cdot 16 = 16$.

3. Final Answer

2.

1. By the Intermediate Value Theorem, there exists a root in the interval $(0, 1)$.

2.

2. $\lim_{x \to 10} (3 - \frac{4}{5}x) = -5$.

2.

3. $\lim_{x \to -1^-} \frac{5}{(x+1)^3} = -\infty$.

2.

4. 16