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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*)