Back to the class
Section 3.5 #45: Find the limit $$\displaystyle\lim_{x \rightarrow -\infty} x+\sqrt{x^2+3}.$$
Solution: Write the function as $$\dfrac{x+\sqrt{x^2+3}}{1}.$$ Multiply top and bottom by $x-\sqrt{x^2+3}$ to get $$\dfrac{x+\sqrt{x^2+3}}{1}=\dfrac{(x+\sqrt{x^2+3})(x-\sqrt{x^2+3})}{x-\sqrt{x^2+3}}=\dfrac{x^2-(x^2+3)}{x-\sqrt{x^2+3}}=\dfrac{-3}{x-\sqrt{x^2+3}}.$$ The highest power of $x$ that appears in the function is $1$ (the $x^2$ is inside a square root so we think of it as having power $1$) so we multiply top and bottom by $\dfrac{1}{x}$ to get $$\dfrac{-\frac{3}{x}}{1-\sqrt{1+\frac{3}{x^2}}}.$$ Now it would be natural to compute $$\displaystyle\lim_{x \rightarrow -\infty} x+\sqrt{x^2+3} =\displaystyle\lim_{x \rightarrow -\infty} \dfrac{-\frac{3}{x}}{1-\sqrt{1+\frac{3}{x^2}}}=\dfrac{0}{1-\sqrt{1+0}},$$ but $1-\sqrt{1}=0$ and so we have a division by zero. What happened? The middle step is invalid because as $x \rightarrow \infty$ we actually have $x$ as negative so in fact the correct middle step should have been $$\displaystyle\lim_{x \rightarrow -\infty} \dfrac{-\frac{3}{x}}{1+\sqrt{1+\frac{3}{|x|^2}}}=\dfrac{0}{1+\sqrt{1+0}}=0.$$

The situation can be clarified with a substitution of variable: let $\alpha = -x$ then we see that $\alpha \rightarrow \infty$ as $x \rightarrow -\infty$. Substituting our $\alpha$ into the formula $\dfrac{-3}{x-\sqrt{x^2+3}}$ yields $$x+\sqrt{x^2+3} = \dfrac{-3}{-\alpha-\sqrt{(-\alpha)^2+3}}=\dfrac{3}{\alpha + \sqrt{\alpha^2+3}}.$$ Therefore $$\displaystyle\lim_{x \rightarrow -\infty} x+\sqrt{x^2+3} = \displaystyle\lim_{\alpha \rightarrow \infty} \dfrac{3}{\alpha + \sqrt{\alpha^2+3}} = \displaystyle\lim_{\alpha \rightarrow \infty} \dfrac{\frac{3}{\alpha}}{1+\sqrt{1+\frac{3}{\alpha^2}}}=0.$$

Section 2.1 #89: Determine whether or not the following function is differentiable at $x=2$: $$f(x) = \left\{ \begin{array}{ll} x^2+1, & \quad x \leq 2 \\ 4x-3, & \quad x>2. \end{array} \right.$$ Solution: We must determine whether the derivative $$f'(2)=\displaystyle\lim_{h \rightarrow 0} \dfrac{f(2+h)-f(2)}{h}$$ exists or not. To do that we will consider both the left limit as $h \rightarrow 0^-$ and the right limit as $h \rightarrow 0^+$. First the limit from the left: $$\displaystyle\lim_{h \rightarrow 0^-} \dfrac{f(2+h)-f(2)}{h} = \displaystyle\lim_{h \rightarrow 0^-} \dfrac{[(2+h)^2+1] - (2^2+1)}{h} = \displaystyle\lim_{h \rightarrow 0^-} 4+h = 4,$$ while the limit from the right: $$\displaystyle\lim_{h \rightarrow 0^+} \dfrac{f(2+h)-f(2)}{h} = \displaystyle\lim_{h \rightarrow 0^+} \dfrac{[4(2+h)-3]-[4\cdot 2-3]}{h} = 4.$$ Therefore the limit from the left equals the limit from the right and so the derivative $f'(2)$ exists.
Section 2.2 #52: Find the derivative of the function $$f(x)=\dfrac{2}{\sqrt[3]{x}}+3\cos(x).$$ Solution: Rewrite as $f(x)=2x^{-\frac{1}{3}}+3\cos(x)$. Now apply the standard rules of differentiation to compute $$\dfrac{\mathrm{d}}{\mathrm{d}x} \left[ 2x^{-\frac{1}{3}}+3\cos(x) \right] = -\dfrac{2}{3} x^{-\frac{4}{3}}-3\sin(x).$$