# Calculus HW2
1.
(1)
$$\lim_{x\to 1}\frac{\tan(x-1)}{\sqrt{x}-1}=\lim_{x\to 1}\frac{\tan(x-1)}{x-1}\times (\sqrt{x}+1)\\ \because\ \lim_{x\to 1}\frac{\tan(x-1)}{x-1}=1\ \left(\lim_{t\to 0}\frac{\tan(t)}{t}=1\right)\\ \therefore\ \lim_{x\to 1}\frac{\tan(x-1)}{\sqrt{x}-1}=\left(\lim_{x\to 1}\frac{\tan(x-1)}{x-1}\right)\times\left(\lim_{x\to 1}\sqrt{x}+1\right)=2\ \blacksquare$$
(2)
$$\lim_{x\to 0}\frac{\tan(5x)}{\sin(4x)}=\lim_{x\to 0}\frac{\tan(5x)}{5x}\times\frac{4x}{\sin(4x)}\times\frac{5}{4}=\frac{5}{4}\ \blacksquare$$
(3)
$$\lim_{x\to 0}\frac{3^{x}-1}{6^{x}-1}=\lim_{x\to 0}\frac{3^{x}-1}{x-0}\cdot\lim_{x\to 0}\frac{1}{\left(\frac{6^{x}-1}{x-0}\right)}=\frac{\frac{d}{dx}(3^{x})\Bigg|_{x=0}}{\frac{d}{dx}(6^{x})\Bigg|_{x=0}}=\frac{\ln(3)}{\ln(6)}\ \blacksquare$$
2.
Hint: Consider the derivatives of some functions and the definition of natural exponential.
(1)
$$\lim_{x \to 0}\frac{e^{x}-1}{x}=\lim_{x \to 0}\frac{e^{x}-1}{x-0}=\frac{d}{dx}(e^{x})\Bigg|_{x=0}=1\ \text{(by definition)}\ \blacksquare$$
(2)
$$\lim_{x\to 0}\frac{\ln(3x+1)}{x}=\ln\left(\lim_{x\to 0}(3x+1)^{\frac{1}{x}}\right)=\ln\left(\lim_{t\to \infty}(1+\frac{3}{t})^{t}\right)=\ln\left(e^{3}\right)=3\ \blacksquare$$
(3)$$\lim_{x\to \frac{\pi}{2}}\frac{1-\sin(x)}{\pi-2x}=2\times\lim_{x\to \frac{\pi}{2}}\frac{\sin(x)-1}{x-\frac{\pi}{2}}=0\ \text{(by definition)}\ \blacksquare$$
3.
(1) Prove by induction:
$$\frac{d}{dx}\sin(x)=\cos(x)=\sin\left(x+\frac{\pi}{2}\right)\\
\because\ \frac{d}{dx}\sin\left(x+\frac{n\pi}{2}\right)=\cos\left(x+\frac{\pi}{2}\right)=\sin\left(x+\frac{(n+1)\pi}{2}\right)\ \text{(by chain rule)}\\
\therefore\ \frac{d^{n}}{dx^{n}}\sin(x)=\sin\left(x+\frac{n\pi}{2}\right)\ \blacksquare$$
(2)
$$\vert x\vert =\sqrt{x^{2}}\Rightarrow\frac{d}{dx}\vert x\vert=\frac{d}{dx}\sqrt{x^{2}}=\frac{1}{2\sqrt{x^{2}}}\cdot 2x=\frac{x}{\vert x\vert}\ \text{(by chain rule)}\ \blacksquare$$
(3)
$$\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\\
\frac{dy}{dx}=2-e^{x},\ \frac{dx}{dt}=\frac{1}{2\sqrt{\ln(t)}}\cdot\frac{1}{t}=\frac{1}{2t\sqrt{\ln(t)}}\ \text{(by chain rule)}\\
\Rightarrow \frac{dy}{dt}=\frac{2-e^{\sqrt{\ln(t)}}}{2t\sqrt{\ln(t)}}\ \blacksquare$$
4.
$$f(x)=\left\{ \begin{array}{ll}
x\sin\left(\frac{1}{x}\right)&,\text{if }x\neq 0\\
0&,\text{if }x=0
\end{array} \right.$$
(1) Continuity of $f(x)$:
$$\forall\ \epsilon >0,\ \exists\ \delta=\epsilon >0\ such\ that\ \vert x-0\vert <\delta\Rightarrow\Bigg\vert x\sin\left(\frac{1}{x}\right)-0\Bigg\vert < \vert x\vert < \delta = \epsilon\\
Hence,\ f(x)\ is\ continuous\ at\ x=0\ \blacksquare$$
(2) Differentiability of $f(x)$:
$$For\ \epsilon =\frac{1}{2}, \forall\ \delta >0,\ \exists\ N\in\mathbb{N}\ such\ that\ x=\frac{1}{\left(N+\frac{1}{2}\right)\pi}\Rightarrow\vert x-0\vert <\delta\\
But\ \Bigg\vert \frac{x\sin\left(\frac{1}{x}\right)-0}{x-0}\Bigg\vert=\Bigg\vert \sin\left[\left(N+\frac{1}{2}\right)\pi\right]\Bigg\vert= 1 > \epsilon=\frac{1}{2}\\
Hence,\ f(x)\ is\ not\ differentiable\ at\ x=0\ \blacksquare$$