Math 242: Calculus I |
Part I, Computations -- Thursday, 18 October 2018
1.
Compute the limits, provided they exist.
a.) | \(\displaystyle \lim_{x \to 0} \frac{x^2 - 2x + 1}{x - 1} \) | b.) | \(\displaystyle \lim_{x \to 1} \frac{x^2 - 2x + 1}{x - 1} \) | c.) |
\(\displaystyle \lim_{\theta \to \tfrac{\pi}{2}^-} \tan\theta \)
|
d.) | \(\displaystyle \lim_{x \to 0} \dfrac{\sin(4x)}{x} \) | e.) | \(\displaystyle \lim_{y \to \tfrac{\pi}{4}} \ln(\tan(y))\) | f.) |
\(\displaystyle \lim_{u \to \frac{\pi}{12}} \sin(u)\cos(u) \)
|
a.) | \( f(x) = 3x^2 + 9x - 5\) | b.) | \( g(x) = \dfrac{\sqrt{x} - \sqrt[3]{x}}{\sqrt[5]{x}}\) | c.) |
\(h(x) = x^2\sin(x + 1) \)
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d.) | \(v(\theta) = \tan\theta\cos\theta \) | e.) | \( y = \tan(\cos(x)) \) | f.) |
\( K(p) = \sec^2(p) \)
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g.) | \( P(T) = \dfrac{nRT}{V} \) | h.) | \( q(x) = -\dfrac{x}{\sqrt{1 + x^2}} \) | i.) |
\( d(x) = \sqrt{1 - (x-1)^2} \)
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j.) | \( f(r) = (2r - 1)^3(r + 1)^2 \) | k.) | \( T(t) = \dfrac{2 - \tan(t)}{\cos^2(t)} \) | l.) |
\( u(y) = \csc(\sqrt{y^3 - 2y^2 + y}) \)
|
Part II, Proofs and Applications -- Friday, 19 October 2018
6. You wish to prove that \(\displaystyle \lim_{x \to 2} 12x - 4 = 20\).
If you fix \(\varepsilon > 0\), what should you set \(\delta\) equal to in order
to finish the proof?
7. Show that the equation \(x^4 - 6x^2 = -5\) has a real root in
the interval \((0,2)\).
8. Verify that the function satisfies the hypotheses of the Mean Value
Theorem on the interval. Then find the values of \(c\) that are guaranteed by
the theorem.
\[
f(x) = x^3 - 3x^2 + x - 1,\ \ \ 0 \leq x \leq 2
\]
9. Use the limit definition of derivative to compute \(\tfrac{dy}{dx}\)
for the given functions. You must use the limit definition to receive credit on
the exam.
a.) |
\(y = \sqrt{x}\) |
|
b.) |
\(y = \dfrac{1}{x}\) |
|
c.) |
\(y = x^5\) |
a.) |
\(y = x^3 - 3x^2 - 9x + 27\) |
|
b.) |
\(y = \dfrac{\sqrt{x^2 - 1}}{x - 1}\) |
|
c.) |
\(y = \dfrac{x^3}{x^2 + 3x + 2}\) |
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