Math 3A Summer 2011, Homework 1


Due June 23 In your TA section .

  1. Compute the limits of the following sequences using the limit laws and any limits we already know (e.g. $\lim_{n\rightarrow\infty}\frac{1}{n}=0$):
    1. $\frac{n}{n+1}+\frac{1}{n^{2}}$
      Answer: \begin{align*} \lim_{n\rightarrow\infty} & \left(\frac{n}{n+1}+\frac{1}{n^{2}}\right) =\lim_{n\rightarrow\infty}\frac{n}{n+1}+\lim_{n\rightarrow\infty}\frac{1}{n^{2}}\\ & =\lim_{n\rightarrow\infty}\frac{1}{1+\frac{1}{n}}+\left(\lim_{n\rightarrow\infty}\frac{1}{n}\right)\left(\lim_{n\rightarrow\infty}\frac{1}{n}\right) \\ & =\frac{\lim_{n\rightarrow\infty} 1}{\lim_{n\rightarrow\infty}(1+\frac{1}{n})} +0\cdot 0 \\ & = \frac{1}{\lim_{n\rightarrow\infty}1+\lim_{n\rightarrow\infty}\frac{1}{n}}=\frac{1}{1}=1\end{align*}
    2. $\frac{4n^{2}-3}{3n^{2}}$
    3. $\frac{n^{2}+2^{-n}}{2n^{2}}$
      4. Answer: \begin{align*} \lim_{n\rightarrow\infty} \frac{n^{2}+2^{-n}}{2n^{2}} & =\lim_{n\rightarrow\infty}\left(\underbrace{\frac{n^{2}}{2n^{2}}}_{=\frac{1}{2}}+\frac{2^{-n}}{2n^{2}}\right) =\lim_{n\rightarrow\infty}\frac{1}{2}+\lim_{n\rightarrow\infty}(2^{-n}\cdot \frac{1}{2}\cdot \frac{1}{n^{2}} ) \\ & =\frac{1}{2}+(\lim_{n\rightarrow\infty}2^{-n})(\lim_{n\rightarrow\infty} \frac{1}{2})(\lim_{n\rightarrow\infty}\frac{1}{n^{2}}) =\frac{1}{2}+0\cdot \frac{1}{2}\cdot 0 = \frac{1}{2}\end{align*}
    4. $\left(\frac{1}{n^{3}}+3\right)(2+2^{-n})$
    5. $\frac{2^{n}+2^{-n}}{2^{n}-2^{-n}}$

  2. Evaluate the following limits using any limit law we've covered:
    1. $\lim_{x\rightarrow 1} \frac{x^{2}-1}{x-1}$.
    2. $\lim_{x\rightarrow 0} \frac{e^{2x}-1}{e^{x}-1}$ (Hint: think of what you did in the previous problem)
    3. $\lim_{x\rightarrow 0} \frac{x^{-2}}{x^{-2}+1}$
    4. $\lim_{x\rightarrow 0} \frac{\sqrt{x^{2}+4}-2}{x}$
      Answer: \begin{align*} \lim_{x\rightarrow 0}\frac{\sqrt{x^{2}+4}-2}{x} & =\lim_{x\rightarrow 0}\frac{\sqrt{x^{2}+4}-2}{x} \frac{\sqrt{x^{2}+4}+2}{\sqrt{x^{2}+4}+2} =\lim_{x\rightarrow 0}\frac{(x^{2}+4)-4}{x(\sqrt{x^{2}+4}+2)} \\ & =\lim_{x\rightarrow 0}\frac{x^{2}}{x(\sqrt{x^{2}+4}+2)} =\lim_{x\rightarrow 0}\frac{x}{\sqrt{x^{2}+4}+2}\\ & =\frac{\lim_{x\rightarrow 0}x}{\lim_{x\rightarrow 0}\sqrt{x^{2}+4}+2} =\frac{0}{\sqrt{0+4}+2}=0\end{align*}
    5. $\lim_{x\rightarrow -5}\frac{x+5}{x^{2}+x-20}$

  3. Let $h(x)$ be the function that rounds down numbers, e.g. $h(1.5)=1$, $h(2)=2$, $h(\pi)=3$.
    1. Compute $\lim_{x\rightarrow 10^{-}}h(x)$ and $\lim_{x\rightarrow 10^{+}}h(x)$.
      Answer Note that if $x$ is close and to the right of the value $10$, then $h(x)=10$, hence \[\lim_{x\rightarrow 10^{+}}h(x)=\lim_{x\rightarrow 10^{+}}10=10,\] since taking the right-hand limit of $h(x)$ means we are letting $x$ tend to $10$ from the positive (i.e. the right) side of $10$. For $x$ close to $10$ but on the left of $10$, $h(x)=9$ (e.g. $h(9.991)=9$), thus \[\lim_{x\rightarrow 10^{-}}h(x)=\lim_{x\rightarrow 10^{-}}9=9.\]
    2. Show that $h(x)$ is continuous at $x=\frac{10}{3}$. (Hint: think of a simple formula for $h(x)$ when $x$ is in $(3,4)$).
      Answer: If $x$ is tending to $\frac{10}{3}$, it is eventually between 3 and 4, hence $h(x)=3$ always, thus $\lim_{x\rightarrow \frac{10}{3}}h(x)=\lim_{x\rightarrow \frac{10}{3}}3=3=h(\frac{10}{3})$, thus $h$ is continuous at $x=\frac{10}{3}$.

  4. Show that $f(x)=|x|$ is continuous at every point in $\mathbb{R}$. (Hint: verify that $\lim_{x\rightarrow c}f(x)=f(c)$ in 3 cases: $c>0$, $c<0$, and $c=0$. For the first two cases, use the fact that the functions $g(x)=x$ and $h(x)=-x$ are continuous everywhere; for the last, show that the left and right hand limits agree.)

    5. Answer: If $c>0$, then if $x$ is close to $c$, it is also bigger than zero and hence $|x|=x$, so we have \[\lim_{x\rightarrow c}|x|=\lim_{x\rightarrow c}x=c=|c|.\] Here, we just used the fact that the function $x$ is continuous at every point, so $\lim_{x\rightarrow c}x=c$. Similarly one can show the same thing for $c<0$. For $c=0$, note that \[\lim_{x\rightarrow 0^{-}}|x|=\lim_{x\rightarrow 0^{-}}-x=-0=0\] since we are taking the limit as $x\rightarrow 0$ from the negative side, so $x<0$ and hence $|x|=-x$. Similarly, one can show $\lim_{x\rightarrow 0^{+}}|x|=0$, thus the left and right hand limits coincide and they both equal $f(0)=|0|$. Hence, $\lim_{x\rightarrow 0}|x|=|0|$ and thus $|x|$ is continuous at 0, and we have checked all possible values of $c$, so $|x|$ is continuous everywhere.
  5. Find the values of $x\in\mathbb{R}$ where the following functions are continuous, using any of the rules we have for combining continuous functions.
    1. $f(x)=5x^{2}+15x+10$
    2. $g(x)=\cos\left(\frac{1}{x+1}\right)$
    3. $\log(1-|x|)$
      Answer: $\log y$ is defined only for $y>0$, hence $\log(1-|x|)$ is defined only when $1-|x|>0$, i.e. $|x|<1$. Since $|x|$ is continuous everywhere (by a previous problem), so is $1-|x|$, hence $\log(1-|x|)$ is continuous wherever it is defined, i.e. for $|x|<1$, or in set notation, it is continuous and defined on $(-1,1)=\{x:-1$\tan 2\pi x$

  6. For each function $f$ below and given point $c$, find a value for $f(c)$ so that $f$ is continuous at $c$.
    1. $f(x)=\frac{x^{2}-16}{x+4}$, $c=4$
    2. $f(x)=\frac{x^{2}-3x+2}{x-2}$, $c=2$.
      Answer: Note that \[\lim_{x\rightarrow 2}f(x)=\lim_{x\rightarrow 2}\frac{x^{2}-3x+2}{x-2}=\lim_{x\rightarrow 2}\frac{(x-2)(x-1)}{x-2} =\lim_{x\rightarrow 2}(x-1)=2-1=1,\] thus if we pick $f(x)=1$, we have $\lim_{x\rightarrow 2}f(x)=1=f(1)$, thus $f$ is continuous at $1$.