Homework 2, due June 29 at beginning of class.

  1. Comput the following limits:
    1. $\lim\limits_{x\rightarrow\infty}\frac{e^{2x}}{e^{2x+5}+1}$
    2. $\lim\limits_{x\rightarrow \infty} \frac{x^{3}+2}{x^{2}+1}$
      Answer: \[\lim_{x\rightarrow\infty}\frac{x^{3}+2}{x^{2}+1}=\lim_{x\rightarrow \infty}\frac{x^{3}}{x^{2}}\frac{1+\frac{2}{x^{3}}}{1+\frac{1}{x^{2}}} = \left(\lim_{x\rightarrow\infty}\frac{x^{3}}{x^{2}}\right)\left(\lim_{x\rightarrow\infty}\frac{1+\frac{2}{x^{3}}}{1+\frac{1}{x^{2}}}\right) \] and since the left limit in the product is infinity while the right limit is finite, the product of the limits will be infinity.
    3. $\lim\limits_{x\rightarrow\infty} \frac{2x^{10}+5x+3}{x^{2}+x^{10}+15}$
    4. $\lim\limits_{x\rightarrow\infty} (\sqrt{2x+1}-\sqrt{2x-1})$ ( Hint: you will need to multiply and divide by something, and remember $(a-b)(a+b)=a^2-b^2$) Answer: \begin{align*} \lim_{x\rightarrow\infty}\sqrt{2x+1}-\sqrt{2x-1} & =\lim_{x\rightarrow\infty}(\sqrt{2x+1}-\sqrt{2x-1})\frac{\sqrt{2x+1}+\sqrt{2x-1}}{\sqrt{2x+1}+\sqrt{2x-1}} \\ & =\lim_{x\rightarrow \infty}\frac{2x+1-(2x-1)}{\sqrt{2x+1}+\sqrt{2x-1}} \\ & = \lim_{x\rightarrow\infty}\frac{2}{\sqrt{2x+1}+\sqrt{2x-1}} \end{align*} At this point, it would be ok to say that this limit was zero, since the denominator is going to infinity (i.e. the denominator is getting larger and larger, so one over the denomonator will get smaller and smaller). For more explicit detail, we continue as such: the above is equal to \begin{align*} \lim_{x\rightarrow\infty}\frac{2}{\sqrt{x}(\sqrt{2+\frac{1}{\sqrt{x}}}+\sqrt{2-\frac{1}{\sqrt{x}}})} & =\lim_{x\rightarrow \infty}\frac{1}{\sqrt{x}}\frac{2}{\sqrt{2+\frac{1}{\sqrt{x}}}+\sqrt{2-\frac{1}{\sqrt{x}}}}\\ & = \left(\lim_{x\rightarrow \infty}\frac{1}{\sqrt{x}}\right)\frac{\lim_{x\rightarrow\infty} 2}{\lim_{x\rightarrow \infty}(\sqrt{2+\frac{1}{\sqrt{x}}}+\sqrt{2-\frac{1}{\sqrt{x}}})}\\ & =0\cdot \frac{2}{\sqrt{2+0}+\sqrt{2-0}}=0.\end{align*}
    5. $\lim\limits_{x\rightarrow\infty}\frac{x}{\sqrt{x^{2}+1}}$

  2. Compute the following limits (these all use the squeeze/sandwich theorem):
    1. $\lim\limits_{x\rightarrow\infty} e^{-x}(\cos x+\sin x)$
      Answer: Since $-2\leq \cos x+\sin x\leq 2$, we have \[-2e^{-x}\leq e^{-x}(\cos x+\sin x)\leq 2e^{-x}\] and since $\lim_{x\rightarrow\infty}(-2e^{-x})=\lim_{x\rightarrow\infty}2e^{-x}=0$, by the squeeze theorem, we have that the limit in question converges to zero.
    2. $\lim\limits_{x\rightarrow 0}\frac{x}{1+\sin x}$
      Answer: For this one, squeeze theorem is not actuallyneeded, since the numerator and denominator are both continuous (and the denominator is not zero at $x=0$) we may simply plug in zero in for $x$. To compute this limit.
    3. $\lim\limits_{x\rightarrow \infty} \frac{1}{x+\cos x}$
    4. $\lim\limits_{x\rightarrow \infty}2^{-x+\cos x}$
      Answer: Since $-1\leq \cos x\leq 1$, we have \[-x-1\leq -x+\cos x\leq -x+1\] and hence \[2^{-x-1}\leq 2^{-x-\cos x}\leq 2^{-x+1}\] and since $\lim 2^{-x+1}=2\lim_{x\rightarrow\infty}2^{-x}=2\cdot 0=0$, and similarly for $2^{-x-1}$, we get that the limit is zero.
    5. $\lim\limits_{x\rightarrow\infty} \frac{\cos(\sin(\sqrt{x}))}{x}$

  3. Intermediate Value Theorem:
    1. Without solving, show that $f(x)=x^{3}+5x-2$ has a root in $[-1,1]$.
      Answer: Note that \[f(-1)=-1-5-2=-8<0\] and \[f(1)=1+5-2=4>0,\] hence, since $0$ is inbetween $f(-1)$ and $f(1)$, the intermediate value theorem tells us there is an $x\in[-1,1]$ so that $f(x)=0$, i.e. $x$ is a root for $x^{3}+5x-2$.
    2. Show that for some $x\in [0,1]$, $\cos x=x$ (Hint: Consider $f(x)=\cos x-x$).

  4. Compute the derivatives of the following function using the definition of a derivative (you can use any limits we've covered so far, e.g. $\lim\limits_{h\rightarrow 0}\frac{\sin h}{h}=1$)
    1. $10$
    2. $x^{3}$
    3. $x^{2}+2x+2$
    4. $\frac{1}{\sqrt{x}}$
    5. $\frac{1}{\sin x}$ ( Hint: At some point, you will require the formula $\sin(a)\cos(b)+\sin(b)\cos(a)=\sin(a+b)$)
      Answer: Actually, disregard that hint (this problem will not get graded). \begin{align*} \left(\frac{1}{\sin x}\right)' & = \lim_{h\rightarrow 0}\frac{\frac{1}{\sin (x+h)}-\frac{1}{\sin x}}{h} \\ & =\lim_{h\rightarrow 0}\frac{\sin x - \sin (x+h)}{h(\sin x)(\sin(x+h))} \\ & = \lim_{h\rightarrow 0} \frac{\sin x-\sin (x+h)}{h}\cdot\frac{1}{(\sin x)(\sin (x+h)}\\ & = \left(\lim_{h\rightarrow 0}\frac{\sin x-\sin (x+h)}{h}\right)\left(\lim_{h\rightarrow 0}\frac{1}{(\sin x)\sin(x+h)}\right) \end{align*} Note that the limit on the left is (by definition of the derivative) $-(\sin x)'=-\cos x$, and the limit on the left is continuous when $\sin x\neq 0$ (again, we're only computing these derivatives where they exist), so we may just plug in $h=0$ and we get \[-\cos x\frac{1}{\sin ^{2}x}.\]

  5. Find the derivative at the given point:
    1. $f(x)=3x+2$, $x=5$
    2. $f(x)=5x^{2}+2x$, $x=1$
    3. $f(x)=-2x^{2}+8$, $x=2$
    4. $f(x)=\frac{1}{x}$, $x=1$
      Answer: \begin{align*} f'(1) & =\lim_{h\rightarrow 0}\frac{f(1+h)-f(1)}{h} =\lim_{h\rightarrow 0}\frac{\frac{1}{1+h}-1}{h} = \lim_{h\rightarrow 0} \frac{\frac{1}{1+h}-\frac{1+h}{1+h}}{h} \\ & = \lim_{h\rightarrow 0} \frac{-h}{h(1+h)} = \lim_{h\rightarrow 0} -\frac{1}{1+h}=-1 \end{align*}
  6. Let \[f(x)=\left\{ \begin{array}{cc} x^{2}\sin\frac{1}{x} & x\neq 0 \\ 0 & x=0\end{array}\right\}.\] Compute the derivative of $f(x)$ at $x=0$ (Hint: when computing the derivative using the definition of a derivative, use the squeeze theorem to compute the limit).
    Answer: We compute the derivative: \[ f'(0)=\lim_{h\rightarrow 0} \frac{f(0+h)-f(0)}{h} =\lim_{h\rightarrow 0} \frac{h^{2}\sin \frac{1}{h}}{h}=\lim_{h\rightarrow 0}h\sin \frac{1}{h}.\] So now we need only show that this limit exists to show that $f$ is differntiable at $x=0$. Note that \[|h\sin \frac{1}{h}|\leq |h|\cdot |\sin\frac{1}{h}|\leq |h|\] since $|\sin\frac{1}{h}|\leq 1$ alays. Note that if $|a|
  7. Bonus Riddle: Suppose it takes Jack all of Monday (starting at midnight) to climb a mountain (so by midnight of Tuesday, he is at the top). Suppose also that it takes all of Tuesday (starting at midnight) to return to the base of the mountain. Show that there is a time of day when Jack is at the same spot on Monday and Tuesday (for example, if the time were 3pm, this would mean that he would be at that spot at 3pm Monday when he's on his way up, and at that same spot at 3pm on Tuesday when he's on his way back down).