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Section 3.5, #15: Compute the limits:
a.) $\displaystyle\lim_{x \rightarrow \infty} \dfrac{x^2+2}{x^3-1}$
b.) $\displaystyle\lim_{x \rightarrow \infty} \dfrac{x^2+2}{x^2-1}$
c.) $\displaystyle\lim_{x \rightarrow \infty} \dfrac{x^2+2}{x-1}$
Solution: For a.), the exponent on the bottom (3) is larger than the exponent on the top (2) and so the limit is $0$. (note: can also multiply top and bottom by $\frac{1}{x^3}$ and conclude same thing)
For b.), the exponent on the top and the exponent on the bottom are the same (2). Therefore we use algebra to see $$\dfrac{x^2+2}{x^2-1} = \dfrac{x^2+2}{x^2-1} \dfrac{\frac{1}{x^2}}{\frac{1}{x^2}}=\dfrac{1+\frac{2}{x^2}}{1-\frac{1}{x^2}}.$$ Now we may compute $$\displaystyle\lim_{x \rightarrow \infty} \dfrac{x^2+2}{x^2-1} = \displaystyle\lim_{x \rightarrow \infty} \dfrac{1+\frac{2}{x^2}}{1-\frac{1}{x^2}}=\dfrac{1+0}{1-0} = 1.$$ For c.), note that exponent on the top (2) is larger than the exponent on the bottom (1) so the term "$x^2$" in the numerator dominates and so the limit diverges to $\infty$. (note: can also multiply top and bottom by $\frac{1}{x^2}$ to see the same thing)

Section 3.5, #31: Compute the limit $\displaystyle\lim_{x \rightarrow \infty} \dfrac{\sqrt{x^2-1}}{2x-1}$.
Solution: In this problem, the highest power of $x$ in the top is "$2$" but it is under the square root so we think of it as "$1$". The highest power in the bottom is $1$ and so we do the following algebra: $$\dfrac{\sqrt{x^2-1}}{2x-1} = \dfrac{\sqrt{x^2-1}}{2x-1} \dfrac{\frac{1}{x}}{\frac{1}{x}}=\dfrac{\sqrt{\frac{1}{x^2}(x^2-1)}}{2-\frac{1}{x}}=\dfrac{\sqrt{1-\frac{1}{x^2}}}{2-\frac{1}{x}}.$$ Now we may compute $$\displaystyle\lim_{x \rightarrow \infty} \dfrac{\sqrt{x^2-1}}{2x-1} = \displaystyle\lim_{x \rightarrow \infty} \dfrac{\sqrt{1-\frac{1}{x^2}}}{2-\frac{1}{x}}= \dfrac{\sqrt{1}}{2} = \dfrac{1}{2}.$$

Section 2.1, #17: Compute the derivative of $f(x)=x^2+x-3$ using the limit definition.
Solution: Compute directly: $$\begin{array}{ll} f'(x) &= \displaystyle\lim_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ &= \displaystyle\lim_{h \rightarrow 0} \dfrac{[(x+h)^2+(x+h)-3]-[x^2+x-3]}{h} \\ &=\displaystyle\lim_{h \rightarrow 0} \dfrac{[(x^2+2xh+h^2)+(x+h)-3]-[x^2+x-3]}{h} \\ &=\displaystyle\lim_{h \rightarrow 0} \dfrac{2xh+h^2+h}{h} \\ &= \displaystyle\lim_{h \rightarrow 0} 2x+h+1 \\ &= 2x+1. \end{array}$$

Section 2.1, #25: Find an equation of the tangent line to the function $f(x)=x^2+3$ at $(-1,4)$.
Solution: Calculate the derivative $f'(x)=2x$. We plug the $x$-value of the point into the derivative to get the slope of the tangent line: $$f(-1)=2(-1)=-2.$$ The tangent line is therefore the line with slope $-2$ passing through the point $(-1,4)$. We use the point-slope form of the equation of the line to get $$y-4=-2(x-(-1)),$$ or $$y-4=-2(x+1)$$ or $$y=-2x+2.$$ Other forms of this equation are also acceptable as an answer.

Section 2.2, #23: Compute the derivative of the function $f(x)=\dfrac{1}{x}-3\sin(x)$.
Solution: The easiest way to proceed would be to rewrite $\dfrac{1}{x}$ as $x^{-1}$ and compute $$f'(x)=\dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{1}{x} - 3\sin(x) \right] = \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ x^{-1} \right] - 3 \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \sin(x) \right]=-x^{-2}-3\cos(x).$$ You may also get rid of the negative exponent in the final answer to write $$f'(x)=-\dfrac{1}{x^2}-3\cos(x).$$