Limits and Derivatives: questions with answers

Turning the divisor $5x$ into the matching angle $11x$ costs a factor: $\frac{\sin 11x}{5x} = \frac{\sin 11x}{11x} \cdot \frac{11x}{5x} = \frac{\sin 11x}{11x} \cdot \frac{11}{5}$. So $k$ is the angle's multiple over the divisor's multiple, that way up, and once the first factor has gone to $1$ it is the whole value of the limit.

As $Q$ tends to $P$ the chord $PQ$ turns until it lies along the TANGENT at $P$, and the limit of its slopes is the slope of that tangent: $f'(7) = \tan\psi$, where $\psi$ is the angle the tangent makes with the horizontal. The chord itself is the only line that can actually be drawn through two given points, which is exactly why it gets chosen; but its slope is one of the quantities whose limit is being taken, not the limit. And $8$ is a height on the curve, not a steepness of anything.

The limit is decided by the heights the two branches run at, and by nothing else. When the branches reach $c$ at different heights the two sides approach different values and there is no limit at all. When both branches run at ONE height the two sides agree, so the limit exists and is that height -- even though the filled circle, which records $f(c)$, sits away from it. So a break drawn at $c$ does not by itself kill the limit, and the filled circle never contributes to it. Nor does any arithmetic on the two marked heights: the number midway between them is approached from neither side.

Theorem 6 gives $\frac{d}{dx}x^{n} = nx^{n-1}$, and the constant multiple rides along. The exponent appears TWICE in the answer and the two appearances differ by one: the ORIGINAL exponent $4$ comes down in front, multiplying the $3$ already there, and the power left upstairs is $4$ reduced by one.

With $u = 4x^{ 6 }$ and $v$ the trigonometric factor, $(uv)' = u'v + uv'$ gives $24x^{ 5 }$ times the factor unchanged, PLUS $4x^{ 6 }$ times that factor's own derivative. The two terms are ADDED -- the minus belongs to the quotient rule, not to this one -- and any minus sign that does appear here comes from $\frac{d}{dx}(\cos x) = -\sin x$, which is a different minus altogether.

The quotient rule's numerator collapses to $1 \times m - 7 \times 3 = m - 21$, so $m - 21 = -16$ and $m = -16 + 21 = 5$. Writing the two terms the other way round would give $21 - m$ instead, and solving that leads to $37$ -- a number of the same size on the wrong side of $21$.

With a letter for the point, all four candidate forms look equally reasonable, which is exactly why the question is worth asking this way: nothing can be checked by arithmetic. Theorem 2 gives $8a^{ 8 - 1}$ -- the exponent comes down in front AND the power drops by one. Leaving the power at $8$ makes only the first change; writing $a^{ 8 - 1}$ with nothing in front makes only the second.

The limit is decided by the heights the graph takes at inputs NEAR $c$, and on both sides those heights are the level the two branches run at, which is the level the unfilled circle sits on. The filled circle records $f(c)$, the value the definition assigns at the point itself, and the limit never uses it. The two marks lie above the same input but they answer different questions.

Taking $g$ to be the constant function $3$ in the product rule gives $\lim [3 f(x)] = 3 \lim f(x) = 3 \times 9$. The constant is unaffected by the limiting process in the sense that it comes OUT of the limit unchanged and still multiplies the answer -- it is neither dropped nor turned into an added term.

The numerator of the difference quotient is $\frac{ 5 }{x + h + 4} - \frac{ 5 }{x + 4}$, and the two fractions must be put over the common denominator $(x + h + 4)(x + 4)$ before anything can be cancelled. That produces $\frac{-5h}{(x + h + 4)(x + 4)}$, whose $h$ cancels the $h$ underneath; letting $h \rightarrow 0$ then leaves $-\frac{ 5 }{(x + 4)^{2}}$. The minus sign is produced by that combination, so any route which skips it loses the sign -- and stopping at the cancellation leaves an answer still containing $h$, which is not a derivative at all.

Once $\pi$ is recognised as a constant, $\pi r^2 + 2$ is a polynomial in $r$, so the limit is its value at $6$: $\pi \times 36 + 2$, written $36\pi + 2$. $\pi$ is not a second name for the variable, and $\frac{22}{7}$ is a rational number while $\pi$ is not, so replacing one by the other changes the value. Only the form that still carries $\pi$ is exact.

Both denominators vanish at $3$, so the two look alike until the numerators are checked. $P$'s numerator also vanishes there and factorises as $(x - 3)(x + 3)$, so the common factor cancels and the limit is $6$. $Q$'s numerator is $14$ at $3$, which is not zero, so nothing cancels and $Q$ has no limit there. A vanishing denominator by itself decides nothing, and it certainly does not make either limit $0$.

Theorem 4 concludes only when the upper and lower limits are the SAME number. Here they are $5$ and $10$, which differ, so the trapping gives an interval and no single value: all that follows is $5 \leq \lim g \leq 10$. The trapping picture looks conclusive on its own, which is why one of the two outer values gets named anyway. And the brackets are not strict: only $\leq$ survives the passage to the limit, however strict the inequalities between the functions are.

Cancelling matching factors leaves $\frac{g(x)}{(x-a)^{2}h(x)}$. The numerator approaches $8$, while the denominator approaches zero. The quotient is unbounded, so the limit does not exist.

The constant comes out of the limit unchanged, so the whole computation turns on which of the two standard values is being multiplied. $\frac{\sin x}{x}$ tends to $1$, so its multiple tends to $6$; $\frac{1 - \cos x}{x}$ tends to $0$, and every multiple of $0$ is $0$, whatever the constant is.