Differential and Integral Calculus (21-120) — Feedback on Homework 2

Homework 2 was due on Thursday 5th September 2013 and consisted of:

• Section 2.2 Q 8, 12, 32, 34, 46
• Section 2.3 Q 2, 18
• Section 2.4 Q 4, 20

The questions I marked were 2.2/12 (out of 2), 2.2/34 (out of 2), 2.3/18 (out of 2) and 2.4/20 (out of 3). Everyone got 1 free point for submitting their homework.

Section 2.2 Q12. Common errors included:

• Not sketching the right functions. For example, a lot of people sketched $1-\sin x$ when $x \lt 0$, rather than $1+\sin x$.
• A lot of people wrote things like $a = (-\infty, \pi) \cup (\pi, \infty)$ to denote the fact that $\lim_{x \to a} f(x)$ exists for all $a$ except for $\pi$. I gave the point for this, but it's bad notation: $(-\infty, \pi) \cup (\pi, \infty)$ is a set, whereas $a$ is a number. To denote the fact that $a$ lies in the set you should write $\in$ instead of $=$. For instance, $x \in (-3, 7)$ means that $x$ lies between $-3$ and $7$ (exclusive). Something like $a \in (-\infty, \pi) \cup (\pi, \infty)$, or even simply $a \ne \pi$, would have been better.
• A few people said that the function was not continuous at $0$, even though the graph joins up there. In fact the function is continuous there, since $$\lim_{x \to 0^-} (1+\sin x) = 1 \quad \text{and} \quad \lim_{x \to 0^+} \cos x = 1$$

Section 2.2 Q34. The main reason why people didn't get both points on this question was failing to justify the answer. If you wrote $\lim_{x \to \pi^-} \cot x = -\infty$ then you'll have got one point. The second point was reserved for justification. I'd have been happy with a graph, or an argument along the lines that, as $x$ approaches $\pi$ from below, $\cos x$ approaches $-1$ and $\sin x$ approaches $0$ from above. Since dividing $-1$ by a small positive number gives a large negative number, the limit is $-\infty$. (Note that as $x \to \pi^+$ the opposite happens: you divide by a small negative number, and so $\lim_{x \to \pi^+} \cot x = +\infty$.)

Section 2.3 Q18. There were two common mistakes on this question:

• Declaring that the limit doesn't exist because the denominator is zero. Sure, if your limit looks like $\frac{a}{0}$ when $a \ne 0$ then the limit blows up. However, when the limit looks like $\frac{0}{0}$ you have to do more work. It might have a limit or it might not: for instance, $\lim_{h \to 0} \frac{h}{h} = 1$, but $\lim_{h \to 0} \frac{h}{h^2}$ does not exist, despite the fact that both limits look like $\frac{0}{0}$. (Incidentally, this is why it's impossible to assign a value to $\frac{0}{0}$ which makes sense.)
• A lot of people either made errors when simplifying the expression. The best way to do this question was to expand $$(2+h)^3 = 8 + 12h + 6h^2 + h^3$$ Then the $8$ cancels with the other $-8$ on the numerator, and the fraction simplifies nicely by cancelling one factor of $h$. Some people tried to do this but didn't manage, but some tried some really exotic tricks, like multiplying by $\frac{(2+h)^3+8}{(2+h)^3+8}$. This is a sure way to make your life difficult! Only do things like this when you're trying to get rid of roots, but not powers.

Section 2.4 Q20. This question was bound to be a challenge! Of the people who found a correct value of $\delta$ (anything in the region $0 \lt \delta \le \frac{5}{4}\varepsilon$ was acceptable), the most common issue was not proving that the value of $\delta$ worked. Partial credit was given for working which was readable and had reversible steps. Other issues included:

• Obtaining the answer $\delta = {-\frac{5}{4}\varepsilon}$. Alarm bells should have been ringing here: the definition of a limit has $\varepsilon \gt 0$ and $\delta \gt 0$ in it: they're both positive, so if your value of $\delta$ is negative, it can't be right. This error usually came from taking a factor of $-1$ outside of modulus brackets. Note that $$\left| \frac{-5}{4}(x-10) \right| = \left| \frac{-5}{4} \right| \left| x-10 \right| = \left( \frac{5}{4} \right) \left| x-10 \right| = \frac{5}{4} \left| x-10 \right|$$ because $-\frac{5}{4} \lt 0$. The general rule is $$|ab| = |a||b|$$ So if you ever take a constant out of modulus brackets, make sure you take its positive part so that you're not accidentally flipping the sign of your expression.
• Unclear working was a big problem. The reason we prove stuff is to convince ourselves (and whoever reads our work) that our reasoning is correct. A large number of homework papers I had contained pages filled with expressions that looked like $-1 \lt |4x-40| \lt 0.2295$, but with no clear link between any two of them. I couldn't give any credit for stuff like this! If you don't understand something, the best thing you can do is come and see me and we'll talk about it. Failing that, just write down your thoughts and lay out your work neatly with clear steps, and chances are you'll get partial credit for having the right idea.

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