The problem asks us to find several limits and analyze the continuity of functions. We will tackle each sub-problem separately. Question 1 deals with evaluating limits. Question 2 deals with evaluating limits of piecewise functions, and then checking for continuity. Question 3 concerns finding a constant to make a piecewise function continuous. Question 4 asks for values of parameters to make a piecewise function continuous. Question 5 asks us to show that there is a root to a given equation in a given interval using the intermediate value theorem (although the function and interval are missing from the image). We will solve (a) from Question 1, and (a) from Question 2. 1. Problem Description Question 1 (a) asks to find the limit $\lim_{x \to 0} \frac{\cos^4 x}{5+2x}$. Question 2 (a) asks to find the limit $\lim_{x \to -3} (2x + |x-3|)$.

1. Problem Description

The problem asks us to find several limits and analyze the continuity of functions. We will tackle each sub-problem separately.

Question 1 deals with evaluating limits.

Question 2 deals with evaluating limits of piecewise functions, and then checking for continuity.

Question 3 concerns finding a constant to make a piecewise function continuous.

Question 4 asks for values of parameters to make a piecewise function continuous.

Question 5 asks us to show that there is a root to a given equation in a given interval using the intermediate value theorem (although the function and interval are missing from the image).

We will solve (a) from Question 1, and (a) from Question

2.

1. Problem Description

Question 1 (a) asks to find the limit $\lim_{x \to 0} \frac{\cos^4 x}{5+2x}$.

Question 2 (a) asks to find the limit $\lim_{x \to -3} (2x + |x-3|)$.

2. Solution Steps

Question 1 (a):

We can evaluate the limit by direct substitution since the function is continuous at $x=0$.

$\lim_{x \to 0} \frac{\cos^4 x}{5+2x} = \frac{\cos^4(0)}{5 + 2(0)} = \frac{1^4}{5+0} = \frac{1}{5}$

Question 2 (a):

We can evaluate the limit by direct substitution since the function is continuous at $x=-3$.

When $x$ is near $-3$, $x-3$ is negative. So $|x-3| = -(x-3) = 3-x$.

Thus, $\lim_{x \to -3} (2x + |x-3|) = \lim_{x \to -3} (2x + (3-x)) = \lim_{x \to -3} (x+3) = -3+3 = 0$.

3. Final Answer

Question 1 (a): $\frac{1}{5}$

Question 2 (a): $0$