Date modified: Fri 2023-06-23 . 11:35 AM
Date created: Sat 2023-10-07 . 09:02 PM
# Week 1 Thursday Problems Reading: Finish reading chapter 6.1, 6.2$\ast$ and $6.3\ast$, 6.4$\ast$, 6.7, 6.8. Preview 6.6 and 7.1. ## Problems. 1. In class we showed Bernoulli's compound interest limit $$ L = \lim_{N\to\infty}\left( 1+ \frac{1}{N} \right)^N $$equals to $e$ by plugging $L$ above into $\ln(x)$ and showed that it equals to $1$, whence $\ln(L)=1$ implying $L=e$. Recall we used continuity of the natural log and L'Hospital's rule. Let us generalized this, mimic the same method and: - Show the following limit $$ R=\lim_{N\to\infty}\left( 1+ \frac{x}{N} \right)^N $$ is actually $e^x$ by plugging $R$ above into the natural log function, and show $\ln(R)=x$, in which case we can conclude $R=e^x$. - Remark. In doing so, you have just demonstrated an important equality $$ e^x = \lim_{N\to\infty}\left( 1+\frac{x}{N} \right)^N $$which is an equivalent **alternate definition** of $e^x$! (Although if we start with this as the definition of $\exp(x)$, it might be a bit of a challenge to obtain its inverse the natural log from it...) 2. Taking natural log first and exploiting its continuity is a common "manipulation" to deal with exotic indeterminant forms that involves exponentiations. In each of the following limit $L$, (1) classify what kind of exotic indeterminant type it is, and (2) find what $L$ is by taking the natural log, $\ln(L)$, determine the limit possibly using L'Hospital's rule, and deducing what $L$ should be from $\ln(L)$. 1. $\displaystyle L=\lim_{x\to0^+} x^x$ 2. $\displaystyle L=\lim_{x\to0^+} x^{\sqrt{x}}$ 3. $\displaystyle L = \lim_{x\to +\infty} x^{1 / x}$ 4. $\displaystyle L = \lim_{x\to +\infty} x^{(e^{-x})}$ 5. $\displaystyle L=\lim_{x\to 0} (\cos x)^{1 / x^2}$ 3. At the very end of class we briefly discussed whether $0^\infty$ is an indeterminant type or not. It is not and the limit should be zero. Let us show this. Suppose $f$ is a positive and that $\lim_{x\to a}f(x)=0$, and $g$ is such that $\lim_{x\to a}g(x)=\infty$, show $$ \lim_{x\to a} f(x)^{g(x)}=0 $$ by putting the limit in to natural log first (like the previous problem), and see what you get. 4. In calculus 1 there was an important limit $$ \lim_{x\to 0} \frac{\sin(x)}{x}=1 $$ Is this an indeterminant form? Of what type? Show the limit is $1$ by L'Hospital's rule. 5. Solve for all possible values of $x$ for each of the following: 1. $e^{3x+2} = 7$ 2. $e^{3x+2} = 12 e^{x-5}$ 3. $e^x + e^{2x} =3$ (Hint: This is secretly a quadratic equation...) 4. $e^x + e^{-x}=5$ (Hint: This is also secretly a quadratic, first multiply by $e^x$ to both sides) 5. $\sinh(x)=7$. 6. $\cosh(x)=7$ 6. In class we showed the hyperbolic functions $\displaystyle\cosh(x)= \frac{e^x+e^{-x}}{2}$ and $\displaystyle\sinh(x)= \frac{e^x-e^{-x}}{2}$ satisfy the **hyperbolic relation**:$$ \cosh^2(x) - \sinh^2(x) =1 $$ Show this is indeed true by plugging them into the LHS and expanding, and you get $1$ at the end. 7. In class we showed, for positive $a > 0$, $\displaystyle\frac{d}{dx}(a^x) = a^x \cdot \ln(a)$. 1. So then what is $$ \int a^x dx =? $$ 2. Find $\displaystyle\lim_{x\to 0} \frac{7^x - 5^x}{x}=?$ 8. Find the following indefinite integrals. 1. $\displaystyle \int e^x\sqrt{1+e^x}\,dx$ 2. $\displaystyle\int \frac{1}{e^{7x}}dx$ 3. $\displaystyle\int \frac{e^{1 / x}}{x^2} \,dx$ 4. $\displaystyle \int (x^e + e^x)dx$ 5. $\displaystyle\int \frac{(1+e^x)^2}{e^x}\,dx$ 6. $\displaystyle \int \frac{\sqrt{1+e^{-x}}}{e^x}\,dx$ 7. $\displaystyle\int \frac{\cos(\ln(x))}{x}\,dx$ 8. $\displaystyle\int \frac{dx}{x\ln x}$ 9. $\displaystyle \int \frac{1}{8-7x}dx$ 10. $\displaystyle\int \tan(x)dx$. Hint: $\tan(x)=\frac{\sin(x)}{\cos(x)}$, and use $u$-substitution somehow. 11. $\displaystyle\int\tanh(x)dx$. Hint: $\tanh(x)=\frac{\sinh(x)}{\cosh(x)}$.... 9. Sometimes you have to apply L'Hospital's rule several time, if after the first time you get another indeterminant form. The hope is eventually you get something you can evaluate. Compute the following limit by several applications of L'Hospital rule. 1. $\displaystyle\lim_{x\to\infty} \frac{2+x+4x^3}{5+x^2+x^3}$ 2. $\displaystyle\lim_{x\to\infty} \frac{(\ln(x))^3}{x^2}$ 10. Here is a situation of "L'Hospital to nowhere". Consider the limit $$ \lim_{x\to +\infty} \frac{\sinh(x)}{\cosh(x)} $$ 1. What indeterminant type is this? 2. Apply "L'Hospital" once, what new indeterminant type do you get? 3. What if you apply "L'Hospital" once more? Does it get anywhere? 4. We in fact "showed" this in class as the limit of $\tanh(x)$ as $x\to\infty$. The proper way to deal with this is to "divide out the largest term". Here, $$ \frac{\sinh(x)}{\cosh(x)}=\tanh(x)=\frac{e^x-e^{-x}}{e^x+e^{-x}} $$divide top and bottom by $e^x$ first. Then evaluate the limit as $x\to\infty$ and see what you get. 11. Which of the following are actually indeterminant types of limit? If it is not indeterminant, state its limit.$$ \begin{matrix} \frac{0}{0} & \frac{\infty}{\infty} & \frac{0}{\infty} & \frac{\infty}{0} \\ \infty + \infty & \infty-\infty & \infty\cdot\infty & 0\cdot\infty \\ 0^0 & 0^\infty & \infty^0 & 1^\infty \end{matrix} $$ ///