[cv] [orcid] [github] [twitter] [researchgate] [youtube] [play chess with me]
Back to the class
Problem A: Compute the limit $$\displaystyle\lim_{x \rightarrow 0} x^4 \sin \left( \dfrac{1}{x^2} \right).$$ Solution: Recall that for any real number $\theta$, $-1 \leq \sin(\theta) \leq 1$. Taking $\theta=\dfrac{1}{x^2}$ yields the inequality $$-1 \leq \sin \left( \dfrac{1}{x^2} \right) \leq 1.$$ Multiply this inequality on all sides by $x^4$, and get $$-x^4 \leq x^4 \sin \left( \dfrac{1}{x^2} \right) \leq x^4.$$ Notice that $\displaystyle\lim_{x \rightarrow 0} x^4 = \displaystyle\lim_{x \rightarrow 0} -x^4=0$. Therefore we see by the squeeze theorem that $$\displaystyle\lim_{x \rightarrow 0} x^4 \sin \left( \dfrac{1}{x^2} \right) = 0.$$ Section 1.3, #63: Compute the limit $$\displaystyle\lim_{x \rightarrow 0} \dfrac{\sin(x)}{5x}.$$ Solution: Recall that $\displaystyle\lim_{x \rightarrow 0} \dfrac{\sin(x)}{5x}=1$. Since constants can pull out of limits (pg. 59, Theorem 1.2, #1), we may compute $$\displaystyle\lim_{x \rightarrow 0} \dfrac{\sin(x)}{5x} = \dfrac{1}{5} \displaystyle\lim_{x \rightarrow 0} \dfrac{\sin(x)}{x}=\dfrac{1}{5}.$$ Section 1.4, #62: Find the constant $a$ so that the function $$f(x) = \left\{ \begin{array}{ll} 3x^3, \quad x \leq 1 \\ ax+5, \quad x > 1 \end{array} \right.$$ is continuous on the entire real line.
Solution: Since polynomials are continuous on the whole real line, we have no problem with continuity of the $f$ for $x < 1$ and for $x>1$. The only place we must consider is at $x=1$. We must pick the value $a$ so that $\displaystyle\lim_{x \rightarrow 1} f(x) = f(1)$. By the definition of the piecewise function, $f(1)=3(1)^3=3$. We must pick $a$ so that $$\displaystyle\lim_{x \rightarrow 1^-} f(x) = \displaystyle\lim_{x \rightarrow 1^+} f(x).$$ First compute $$\displaystyle\lim_{x \rightarrow 1^-} f(x) = \displaystyle\lim_{x \rightarrow 1^-} 3x^3 = 3,$$ and then compute $$\displaystyle\lim_{x \rightarrow 1^+} f(x) = \displaystyle\lim_{x \rightarrow 1^+} ax+5=a+5.$$ So we see we must pick $a$ so that $$a+5=3,$$ which when we solve for $a$ gives $a=-2$.

Section 1.4, #87: Explain why the function $f(x)=\dfrac{1}{12} x^4 - x^3 +4$ has a zero in the interval $[1,2]$.
Solution: We know that all polynomials are continuous (pg.75), and so we may apply the intermediate value theorem (pg.77, Theorem 1.13). Calculate $$f(1)=\dfrac{1}{12} 1^4 - 1^3 + 4 = \dfrac{1}{12} + 3=\dfrac{37}{12}$$ and calculate $$f(2)=\dfrac{1}{12} 2^4 - 2^3 + 4 = \dfrac{16}{12} - 4 = \dfrac{16}{12} - \dfrac{48}{12} = -\dfrac{32}{12} = -\dfrac{8}{3}.$$ Therefore since $0$ is between the numbers $\dfrac{37}{12}$ and $\dfrac{-32}{12}$, the intermediate value theorem tells us that for some $c$ between $1$ and $2$, the function must take the value $f(c)=0$, which is a zero of $f$.

Section 1.5, #6: Determine whether $f(x)=\dfrac{-1}{x-4}$ approaches $\infty$ or $-\infty$ as $x$ approaches $4$ from the left or from the right.
Solution: Recall that the graph of $g(x)=\dfrac{1}{x}$ looks like this:

So applying a function transformation we see that the graph of $h(x)=\dfrac{1}{x-4}$ looks like this:
Now we see that $f(x)$ is the negative of $h(x)$, and so it flips positive values to negative and vice versa, giving a graph that looks like this:
Therefore we see that $$\displaystyle\lim_{x \rightarrow 4^-} f(x) = \displaystyle\lim_{x \rightarrow 4^-} \dfrac{-1}{x-4} = \infty$$ and $$\displaystyle\lim_{x \rightarrow 4^+} f(x) = \displaystyle\lim_{x \rightarrow 4^+} \dfrac{-1}{x-4} = -\infty.$$ (note: this problem could also be solved by observing that at a number close to but less than $4$, say $3.999$, yields $f(3.999)$ to be a positive number and at a number close to but greater than $4$, say $4.001$, yields $f(4.001)$ to be a negative number to help reconstructing the necessarily information instead of using function transformations)