Advanced Calculus I - Dr. Block Sample Exam 1 with Solutions Fall 2019

Solution:

We need to show that for every > 0, there exists a > 0 such that if 0 < |x - 2| < , then |(x - 4)/(x - 2) - 4| < .

First, we simplify the expression:

For the epsilon-delta proof:

Let > 0 be given. We need to find such that if 0 < |x - 2| < , then |(x + 2) - 4| < .

This simplifies to |x - 2| < .

So we can choose = .

Now, if 0 < |x - 2| < = , then |(x - 4)/(x - 2) - 4| = |x + 2 - 4| = |x - 2| < .

Therefore, by the - definition, lim(x2) (x - 4)/(x - 2) = 4.

Solution:

First, we note that f(x) = (x - 1)/(x - 4) is a rational function.

Rational functions are continuous everywhere except where the denominator equals zero.

To find where the function is not continuous, we solve for x - 4 = 0:

Therefore, the function is not continuous at x = -2 and x = 2.

Since there are no other restrictions, f(x) is continuous on the intervals:

Solution:

a) lim(x) (3x + 2x - 1)/(5x - 3)

To evaluate this limit, we divide both numerator and denominator by x (the highest power of x in the denominator):

b) lim(x0) (sin(3x))/(sin(5x))

We can use the standard limit lim(t0) (sin(t))/t = 1:

c) lim(x0) (1 + x)^(1/x)

Let y = (1 + x)^(1/x). Then ln(y) = ln(1 + x)/x.

As x 0, ln(1 + x) x (using the Taylor series), so ln(y) x/x = 1.

Therefore, lim(x0) ln(y) = 1, which means lim(x0) y = e^1 = e.

Solution:

We'll use the chain rule: d/dx[ln(u)] = (1/u)du/dx.

Let u = sin(x + 1), then f(x) = ln(u).

First, find du/dx:

Now apply the chain rule:

Solution:

First, find the derivative:

To find critical numbers, set f'(x) = 0:

So the critical numbers are x = -1 and x = 3.

Now, we determine the sign of f'(x) on the intervals (-, -1), (-1, 3), and (3, ):

On (-, -1), pick x = -2: f'(-2) = 3(-2) - 6(-2) - 9 = 12 + 12 - 9 = 15 > 0, so f is increasing on this interval.

On (-1, 3), pick x = 0: f'(0) = 3(0) - 6(0) - 9 = -9 < 0, so f is decreasing on this interval.

On (3, ), pick x = 4: f'(4) = 3(4) - 6(4) - 9 = 48 - 24 - 9 = 15 > 0, so f is increasing on this interval.

Therefore, f is increasing on (-, -1) (3, ) and decreasing on (-1, 3).

Solution:

The Mean Value Theorem states that if f is continuous on [a, b] and differentiable on (a, b), then there exists at least one c in (a, b) such that f'(c) = [f(b) - f(a)]/(b - a).

For a = 0 and b = 2, we first compute f(2) - f(0):

Now, find f'(x):

Set f'(c) = 1 and solve for c:

Since we need c in the interval (0, 2), we select c = 2/3.

Solution:

a) (xe^x) dx

Let u = x, then du = 2x dx or (1/2)du = x dx.

b) (sin(x)cos(x)) dx

Let u = sin(x), then du = cos(x) dx.

c) (xln(x)) dx

We'll use integration by parts: u dv = uv - v du.

Let u = ln(x), dv = x dx.

Using the formula:

Solution:

First, we'll use the double-angle identity: sin(x) = (1 - cos(2x))/2.

Solution:

The Fundamental Theorem of Calculus (FTC) Part 1 tells us that if F(x) = ^x f(t) dt, then F'(x) = f(x).

However, in this case, the upper limit of integration is x, not just x. We need to apply the chain rule.

Let u = x, so F(x) = ^u cos(t) dt.

By FTC, dF/du = cos(u) = cos((x)) = cos(x).

By the chain rule, dF/dx = (dF/du)(du/dx) = cos(x)3x.

Solution:

First, we'll rewrite the general term using partial fraction decomposition:

This is a telescoping series. Let's write out the partial sum S:

Now, take the limit as n :

Therefore, the series converges, and its sum is 1.

Solution:

We'll use the Comparison Test with the series (1/n), which is a convergent p-series with p = 2 > 1.

Compare the terms:

For n 1, we have n + 1 n, so:

This means 0 < a b for all n 1.

Since (1/n) converges (p-series with p = 2 > 1), by the Comparison Test, (n/(n + 1)) also converges.

Solution:

We'll use the Ratio Test to find the radius of convergence.

Let a = (x-1)/(n2). Then:

By the Ratio Test, the series converges when:

So the radius of convergence is R = 2.

The interval of convergence is |x-1| < 2, which gives:

Now, we need to check the endpoints x = -1 and x = 3:

For x = -1, the series becomes (1/n), which is the divergent harmonic series.

For x = 3, the series becomes ((-1)/n), which converges by the Alternating Series Test.

Therefore, the interval of convergence is (-1, 3] (including 3 but not -1).