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3.7. Homework Problems

1. (a) Prove, using the Weierstrass definition of the limit, that if \(\lim_{x\rightarrow a} f(x) = F\) and \(\lim_{x\rightarrow a} g(x) = G\) both exist, them \(\lim_{x\rightarrow a} [f(x)+g(x)] = F+G\), i.e., that the limit of a sum is the sum of the limits. (b) Prove the same thing using the definition of the limit in terms of infinitesimals.

2. Sketch the graph of the function e-1/x, and evaluate the following four limits:

\begin{align} \lim_{x\rightarrow 0^{+}} & e^{-1/x} \\ \lim_{x\rightarrow 0^{-}} & e^{-1/x} \\ \lim_{x\rightarrow +\infty} & e^{-1/x} \\ \lim_{x\rightarrow -\infty} & e^{-1/x} \end{align}

3. Verify the following limits.

\begin{align} \lim_{s\rightarrow 1} \frac{s^3-1}{s-1} = 3 \\ \lim_{\theta\rightarrow 0} \frac{1-\cos\theta}{\theta^2} = \frac{1}{2} \\ \lim_{x\rightarrow \infty} \frac{5x^2-2x}{x} = \infty \\ \lim_{n\rightarrow \infty} \frac{n(n+1)}{(n+2)(n+3)} = 1 \\ \lim_{x\rightarrow \infty} \frac{ax^2+bx+c}{dx^2+ex+f} = \frac{a}{d} \end{align}

[Granville, 1911]

4. Evaluate

\[ \lim_{x\rightarrow 0} \frac{x\cos x}{1-2^x} \]

exactly, and check your result by numerical approximation.

5. Amy is asked to evaluate

\[ \lim_{x\rightarrow 0} \frac{e^x}{x} . \]

She applies l'Hôpital's rule, differentiating top and bottom to find 1/ex, which equals 1 when she plugs in x=0. What is wrong with her reasoning?

6. Evaluate

\[ \lim_{u\rightarrow 0} \frac{u^2}{e^u+e^{-u}-2} \]

exactly, and check your result by numerical approximation.

7. Evaluate

\[ \lim_{t\rightarrow \pi} \frac{\sin t}{t-\pi} \]

exactly, and check your result by numerical approximation.

8. Prove a form of l'Hôpital's rule stating that

\[ \lim_{x\rightarrow\infty}\frac{f(x)}{g(x)} \]

is equal to the limit of f'/g' at infinity. Hint: change to some new variable u such that x→∞ corresponds to u→0.

9. Prove that the linear function y=ax+b, where a and b are real, is continuous, first using the definition of continuity in terms of infinitesimals, and then using the definition in terms of the Weierstrass limit.