108學年度｜微積分模組班｜Class 01｜Homework 2.

Homework 2

• Date: 09/26/108 (Week 03)
• Place: 新生教學樓 301

Problem 1.

• Suppose that $\displaystyle \lim_{x \to a} f(x) = L$ and $\displaystyle \lim_{x \to a} g(x) = M$. Prove that $\displaystyle \lim_{x \to a} f(x) − g(x) = L − M$

Solution.

• Note that by definition of limit, for any $\epsilon >0$, there exist $\delta_1, \delta_2 >0$ such that: $$|f(x) - L| < \dfrac{\epsilon}{2} \text{ when } |x-a|<\delta_1$$ $$|g(x) - M| < \dfrac{\epsilon}{2} \text{ when }|x-a|<\delta_2$$ Let $\delta_0 = \min\{\delta_1, \delta_2\}$, then for $|x-a|<\delta_0$, we have: \begin{align*} |(f(x)-g(x))-(L-M)| &= |(f(x)-L) - (g(x) - M)|\\ &\leqslant |f(x)-L| + |g(x)-M| \\ &< \dfrac{\epsilon}{2} + \dfrac{\epsilon}{2} \end{align*} That is, $\displaystyle \lim_{x \to a} f(x) − g(x) = L − M$.

Problem 2.

• Verify that $f(x) = x^6 + x$ is continuous at $x = 1$ by using the definition of limits.

Solution.

• For any $\epsilon >0$, pick $\delta(\epsilon) = \min\left\{1, \dfrac{\epsilon}{64} \right\}$, note that if $|x-1|<1$, then we have $0 < |x| < 2$. Hence for $|x-1|<\delta$, we have: \begin{align*} |(x^6+x)-(1^6+1)| &= |(x^6-1)+(x-1)| \\ &=|(x-1)(x^5+x^4+x^3+x^2+x+1) + (x-1)| \\ &=|x-1||x^5+x^4+x^3+x^2+x+2| \\ &\leqslant |x-1|\left( |x|^5 + |x|^4 + |x|^3 + |x^2| + |x| +2 \right) \\ &< \dfrac{\epsilon}{64} \cdot (2^5 + 2^4 + 2^3 + 2^2 + 2 + 2) = \epsilon \end{align*} So by definition, $f(x)$ is continuous at $x = 1$.

Problem 3.

• Given a function $f(x)$ and $x_0$, and $\epsilon >0$, find a value $\delta > 0$ such that for $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<\epsilon$ holds.
1. $f(x) = \sqrt{x}, x_0 = \dfrac{1}{4}$ and $\epsilon = 0.1$.
2. $f(x) = x^2, x_0 = -2$ and $\epsilon = 0.1$.

Solution.

1. Let $\delta = \min\left\{\dfrac{1}{4}, \dfrac{\epsilon}{2}\right\} = 0.05$, note that if $\left|x-\dfrac{1}{4}\right|<\dfrac{1}{4}$, we have: \begin{align*} \left|x-\dfrac{1}{4}\right|<\dfrac{1}{4} &\Longleftrightarrow -\dfrac{1}{4} < x - \dfrac{1}{4} < \dfrac{1}{4} \\ &\Longleftrightarrow 0 < x < \dfrac{1}{2} \\ &\Longleftrightarrow 0 < \sqrt{x} < \dfrac{1}{\sqrt{2}} \\  &\Longleftrightarrow \dfrac{1}{2} < \sqrt{x} + \dfrac{1}{2} < \dfrac{1}{\sqrt{2}} + \dfrac{1}{2} \end{align*} Hence for $\left|x - \dfrac{1}{4}\right| < \delta$, we have:  $$\left|\sqrt{x} - \dfrac{1}{2}\right| = \dfrac{\left|x - \dfrac{1}{4}\right|}{\left|\sqrt{x}+\dfrac{1}{2}\right|} < \dfrac{\left|x - \dfrac{1}{4}\right|}{\dfrac{1}{2}} < \dfrac{\dfrac{\epsilon}{2}}{\dfrac{1}{2}} = \epsilon $$
2. Let $\delta = \min\left\{1, \dfrac{\epsilon}{5}\right\} = 0.02$, note that if $\left|x-(-2)\right|<1$, we have: \begin{align*} \left|x-(-2)\right|<1 &\Longleftrightarrow -1 < x + 2 < 1 \\ &\Longleftrightarrow -5 < x - 2 < -3 \\ &\Longleftrightarrow 3 < |x-2| < 5 \end{align*} Hence for $\left|x - (-2)\right| < \delta$, we have: $$\left|x^2 - 4\right| = |x-2||x+2| < 5 \cdot \dfrac{\epsilon}{5} = \epsilon$$