Calculus: limits and continuity; derivatives

Let $f(x)=(bx-a)^n$ with $n$ a positive integer, and write $u=bx-a$. By definition,

$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{(u+bh)^n-u^n}{h}.$

Expand by the binomial theorem:

$(u+bh)^n=u^n+nu^{n-1}(bh)+\binom n2u^{n-2}(bh)^2+\cdots+(bh)^n.$

So

$$f'(x)=\lim_{h\to0}\frac{nbhu^{n-1}+\binom n2b^2h^2u^{n-2}+\cdots+b^nh^n}{h} =\lim_{h\to0}\left[nbu^{n-1}+\binom n2b^2hu^{n-2}+\cdots+b^nh^{n-1}\right].$$

Every term containing $h$ vanishes as $h\to0$, leaving

$\frac{d}{dx}(bx-a)^n=nb\,(bx-a)^{n-1}.$

Criteria. Let $f$ be differentiable at $x=c$.

• Local maximum at $x=c$: $f'(c)=0$ and $f''(c)<0$.
• Local minimum at $x=c$: $f'(c)=0$ and $f''(c)>0$.

(If $f'(c)=0$ and $f''(c)=0$, use the sign change of $f'$ around $c$.)

Extrema of $f(x)=2x^3-15x^2+36x+10$.

$f'(x)=6x^2-30x+36=6(x-2)(x-3),\qquad f''(x)=12x-30.$

Setting $f'(x)=0$ gives $x=2$ or $x=3$.

• At $x=2$: $f''(2)=-6<0$, so a local maximum, with $f(2)=16-60+72+10=38$.
• At $x=3$: $f''(3)=6>0$, so a local minimum, with $f(3)=54-135+108+10=37$.

Local maximum value $=38$ at $x=2$; local minimum value $=37$ at $x=3$.

Point of inflection. $f''(x)=0\Rightarrow x=\dfrac52$. Since $f''$ changes sign there ($f'''(x)=12\ne0$), it is a point of inflection:

$f\!\left(\tfrac52\right)=2\cdot\frac{125}{8}-15\cdot\frac{25}{4}+36\cdot\frac52+10=\frac{125}{4}-\frac{375}{4}+90+10=\frac{75}{2}.$

Point of inflection: $\left(\dfrac52,\dfrac{75}2\right)$.

Let $f(x)=\sqrt{2-3x}$. By the definition,

$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\sqrt{2-3x-3h}-\sqrt{2-3x}}{h}.$

Multiply numerator and denominator by the conjugate $\sqrt{2-3x-3h}+\sqrt{2-3x}$:

$f'(x)=\lim_{h\to0}\frac{(2-3x-3h)-(2-3x)}{h\left(\sqrt{2-3x-3h}+\sqrt{2-3x}\right)}=\lim_{h\to0}\frac{-3h}{h\left(\sqrt{2-3x-3h}+\sqrt{2-3x}\right)}.$

Cancel $h$ and let $h\to0$:

$f'(x)=\frac{-3}{2\sqrt{2-3x}}.$

So $\dfrac{d}{dx}\sqrt{2-3x}=-\dfrac{3}{2\sqrt{2-3x}}$.

Criteria. Let $y=f(x)$ be differentiable at $x=a$.

1. First derivative test: $f'(a)=0$ is necessary. If $f'$ changes sign from $+$ to $-$ as $x$ passes through $a$, there is a local maximum at $a$; if $f'$ changes from $-$ to $+$, there is a local minimum.
2. Second derivative test: if $f'(a)=0$ and $f''(a)<0$, then $f$ has a local maximum at $a$; if $f'(a)=0$ and $f''(a)>0$, then $f$ has a local minimum at $a$. (If $f''(a)=0$ the test fails and the first derivative test is used.)

Application. $f(x)=4x^3-15x^2+12x+7$.

$f'(x)=12x^2-30x+12=6(2x^2-5x+2)=6(2x-1)(x-2).$

Setting $f'(x)=0$ gives $x=\dfrac12$ and $x=2$.

$f''(x)=24x-30.$

At $x=\dfrac12$: $f''=12-30=-18<0$, so there is a local maximum.

$f\!\left(\tfrac12\right)=4\cdot\tfrac18-15\cdot\tfrac14+12\cdot\tfrac12+7=\tfrac12-\tfrac{15}{4}+6+7=\tfrac{39}{4}.$

At $x=2$: $f''=48-30=18>0$, so there is a local minimum.

$f(2)=32-60+24+7=3.$

Hence the local maximum value is $\dfrac{39}{4}$ at $x=\dfrac12$, and the local minimum value is $3$ at $x=2$.

Use $1-\cos\theta=2\sin^2\dfrac{\theta}{2}$:

$$\lim_{x\to0}\frac{1-\cos mx}{1-\cos nx}=\lim_{x\to0}\frac{2\sin^2\frac{mx}{2}}{2\sin^2\frac{nx}{2}} =\lim_{x\to0}\frac{\left(\frac{\sin(mx/2)}{mx/2}\right)^2\cdot\frac{m^2x^2}{4}}{\left(\frac{\sin(nx/2)}{nx/2}\right)^2\cdot\frac{n^2x^2}{4}}.$$

Since $\dfrac{\sin\theta}{\theta}\to1$ as $\theta\to0$, this equals

$\frac{m^2}{n^2}\qquad(n\neq0).$

Here $x=2at$ and $y=2at^2$, so

$\frac{dx}{dt}=2a,\qquad \frac{dy}{dt}=4at.$

Therefore

$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{4at}{2a}=2t.$

Use $2\sin A\sin B=\cos(A-B)-\cos(A+B)$:

$\sin2x\sin3x=\frac12\left[\cos x-\cos5x\right].$

Hence

$\int\sin2x\sin3x\,dx=\frac12\left[\sin x-\frac{\sin5x}{5}\right]+C=\frac12\sin x-\frac1{10}\sin5x+C.$

$f(x)=\dfrac12x^2-x$, so $f'(x)=x-1$.

• $f'(x)>0$ when $x>1$, so $f$ is increasing on $(1,\infty)$.
• $f'(x)<0$ when $x<1$, so $f$ is decreasing on $(-\infty,1)$.

At $x=1$, $f'(1)=0$, which is the turning point (minimum, $f(1)=-\tfrac12$).

Left-hand limit: $\lim_{x\to\frac32^-}f(x)=\lim_{x\to\frac32^-}(3-x)=3-\frac32=\frac32.$

Right-hand limit: $\lim_{x\to\frac32^+}f(x)=\lim_{x\to\frac32^+}(-3+x)=-3+\frac32=-\frac32.$

Value at the point: $f\!\left(\tfrac32\right)=-3+\tfrac32=-\tfrac32$.

Since $\dfrac32\ne-\dfrac32$, the left and right limits differ, so $\lim_{x\to3/2}f(x)$ does not exist. Hence $f(x)$ is not continuous at $x=\dfrac32$; it has a jump discontinuity.

Making it continuous. Because the limit does not exist, the discontinuity cannot be removed by merely redefining the value $f(3/2)$. We must change one of the pieces so that both one-sided limits agree. For example, if the first piece is changed to $-x$ (that is, $k-x$ with $k-\tfrac32=-\tfrac32$, so $k=0$), then the left limit becomes $-\tfrac32$, which equals the right limit and $f\!\left(\tfrac32\right)$, and the function is continuous at $x=\tfrac32$.

The circle $x^2+y^2=36$ has centre $(0,0)$ and radius $6$. It is symmetric about both axes, so

$\text{Area}=4\times(\text{area in the first quadrant})=4\int_0^6\sqrt{36-x^2}\,dx.$

Using $\displaystyle\int\sqrt{a^2-x^2}\,dx=\frac x2\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac xa$ with $a=6$:

$$\text{Area}=4\left[\frac x2\sqrt{36-x^2}+18\sin^{-1}\frac x6\right]_0^6 =4\left[(0+18\cdot\tfrac{\pi}{2})-(0+0)\right]=4\cdot9\pi=36\pi.$$

Area $=36\pi$ square units, which agrees with $\pi r^2=\pi\cdot6^2$.

$\lim_{x\to p}\frac{x^2-p^2}{\tan(x-p)}=\lim_{x\to p}\frac{(x-p)(x+p)}{\tan(x-p)}=\lim_{x\to p}\frac{x-p}{\tan(x-p)}\cdot(x+p)$

Put $x-p=h$, so $h\to0$ and $\dfrac{h}{\tan h}\to1$. Hence the limit is $1\times(p+p)=2p$.

Here $x=2a\tan\theta$ and $y=a\sec\theta$, so

$\frac{dx}{d\theta}=2a\sec^2\theta,\qquad \frac{dy}{d\theta}=a\sec\theta\tan\theta.$

$\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{a\sec\theta\tan\theta}{2a\sec^2\theta}=\frac{\tan\theta}{2\sec\theta}=\frac{\sin\theta}{2}.$

Put $u=3\sin x-4$. Then $du=3\cos x\,dx$, so $\cos x\,dx=\dfrac{du}{3}$.

$\int(3\sin x-4)^2\cos x\,dx=\frac13\int u^2\,du=\frac{u^3}{9}+C=\frac{(3\sin x-4)^3}{9}+C.$

$f(x)=15x^2-14x+1$, so $f'(x)=30x-14$. A function is increasing where $f'>0$ and decreasing where $f'<0$.

At $x=\dfrac25$: $f'\!\left(\dfrac25\right)=30\cdot\dfrac25-14=12-14=-2<0$, so $f$ is decreasing.

At $x=\dfrac52$: $f'\!\left(\dfrac52\right)=30\cdot\dfrac52-14=75-14=61>0$, so $f$ is increasing.

Limit. At $x=2$ the expression is of the form $\dfrac00$. Multiply numerator and denominator by the conjugate $\sqrt x+\sqrt{6-x^2}$:

$\lim_{x\to2}\frac{\sqrt x-\sqrt{6-x^2}}{x-2}=\lim_{x\to2}\frac{x-(6-x^2)}{(x-2)\left(\sqrt x+\sqrt{6-x^2}\right)}=\lim_{x\to2}\frac{x^2+x-6}{(x-2)\left(\sqrt x+\sqrt{6-x^2}\right)}.$

Since $x^2+x-6=(x+3)(x-2)$, cancel $(x-2)$:

$=\lim_{x\to2}\frac{x+3}{\sqrt x+\sqrt{6-x^2}}=\frac{5}{\sqrt2+\sqrt2}=\frac{5}{2\sqrt2}=\frac{5\sqrt2}{4}.$

OR: continuity at $x=1$.

Left limit: $\lim_{x\to1^-}(2x+3)=5$.

Right limit: $\lim_{x\to1^+}(6x-1)=5$.

So $\lim_{x\to1}f(x)=5$, but $f(1)=4\ne5$. The limit exists but does not equal the function value, so $f$ is not continuous at $x=1$.

It can be made continuous by redefining $f(1)=5$ (removable discontinuity).

Intersection. Substitute $y=x$ in $y^2=4x$: $x^2=4x\Rightarrow x(x-4)=0$, so $x=0$ or $x=4$. The points are $(0,0)$ and $(4,4)$.

Which is above. For $0<x<4$, the upper branch of the parabola $y=2\sqrt x$ lies above the line $y=x$ (e.g. at $x=1$: $2>1$).

Area

$A=\int_0^4\left(2\sqrt x-x\right)dx=\left[2\cdot\frac{2}{3}x^{3/2}-\frac{x^2}{2}\right]_0^4=\left[\frac43x^{3/2}-\frac{x^2}{2}\right]_0^4.$

$A=\frac43\cdot8-\frac{16}{2}=\frac{32}{3}-8=\frac83.$

The area is $\dfrac83$ square units.

(c) $\dfrac{-3x^2}{(x^2-2)^2}$ (closest listed option)

By the quotient rule,

$\frac{d}{dx}\left(\frac{3x}{x^2-2}\right)=\frac{3(x^2-2)-3x\cdot2x}{(x^2-2)^2}=\frac{-3x^2-6}{(x^2-2)^2}=\frac{-3(x^2+2)}{(x^2-2)^2}.$

The numerator is $-3x^2-6$; option (c) keeps only the $-3x^2$ term and is the nearest of the printed options, which appear to be mistyped.

(b) Rs.34

MC $=\dfrac{dC}{dQ}=30+Q$. At $Q=4$, MC $=30+4=$ Rs.34.

Definitions of the types of discontinuity (2 marks)

A function $f$ is discontinuous at $x=a$ if it is not continuous there, i.e. if any one of the three conditions of continuity fails. The main types are:

1. Removable discontinuity (discontinuity of the first kind): The limit $\lim_{x\to a} f(x)$ exists but is not equal to $f(a)$, or $f(a)$ is not defined. The gap can be removed by redefining $f(a)$.
   Example: $f(x)=\dfrac{x^2-1}{x-1}$ at $x=1$.

2. Jump discontinuity (discontinuity of the first kind): Both one-sided limits exist and are finite, but they are unequal, so $\lim_{x\to a} f(x)$ does not exist.
   Example: $f(x)=[x]$ (greatest integer function) at every integer.

3. Infinite discontinuity (discontinuity of the second kind): At least one of the one-sided limits is infinite (tends to $\pm\infty$).
   Example: $f(x)=\dfrac{1}{x}$ at $x=0$.

4. Oscillatory discontinuity (discontinuity of the second kind): The function oscillates infinitely and the limit fails to exist, though the values stay bounded.
   Example: $f(x)=\sin\dfrac{1}{x}$ at $x=0$.

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Conditions for increasing, decreasing and concavity (3 marks)

Let $f$ be differentiable on an interval $I$.

Increasing:
$f$ is increasing on $I$ if for $x_1<x_2$ in $I$, $f(x_1)<f(x_2)$.
Condition: $f'(x)>0$ for all $x\in I$.
(If $f'(x)\ge 0$, $f$ is non-decreasing.)

Decreasing:
$f$ is decreasing on $I$ if for $x_1<x_2$ in $I$, $f(x_1)>f(x_2)$.
Condition: $f'(x)<0$ for all $x\in I$.
(If $f'(x)\le 0$, $f$ is non-increasing.)

Concavity:

• $f$ is concave upward (convex) on $I$ if $f''(x)>0$ for all $x\in I$; the curve lies above its tangent.
• $f$ is concave downward (concave) on $I$ if $f''(x)<0$ for all $x\in I$; the curve lies below its tangent.
• A point where the concavity changes (i.e. $f''(x)=0$ or undefined and sign of $f''$ changes) is a point of inflection.

Let
$u=x^3 \quad\Rightarrow\quad du=3x^2\,dx \quad\Rightarrow\quad x^2\,dx=\frac{du}{3}$

Then,

$$\int \frac{x^2\,dx}{\sqrt{a^6-x^6}} =\frac13\int \frac{du}{\sqrt{a^6-u^2}}$$

Using $\displaystyle \int \frac{du}{\sqrt{a^6-u^2}}=\sin^{-1}\frac{u}{a^3}+C$,

$$=\frac13\sin^{-1}\frac{x^3}{a^3}+C$$

Answer:

$$\boxed{\frac13\sin^{-1}\left(\frac{x^3}{a^3}\right)+C}$$

(i)
Original price = New price + decline = $15 + 5 = \text{Rs. }20$
Original demand = New demand − increase = $50 - 20 = 30$ units

$E_d=\frac{\%\text{ change in quantity demanded}}{\%\text{ change in price}}$

$\%\text{ change in demand}=\frac{20}{30}\times 100=66.67\%$

$\%\text{ change in price}=\frac{-5}{20}\times 100=-25\%$

$E_d=\frac{66.67}{-25}=-2.67$

So, elasticity of demand is $-2.67$ (or 2.67 in absolute value).

(ii)
The negative elasticity means that price and quantity demanded move in opposite directions: when price falls, demand rises, and when price rises, demand falls. This confirms the law of demand. Since $|E_d|>1$, demand is relatively elastic.

22. (a)

Derivative vs anti-derivative

Physical meaning. If $s(t)$ is the position of a body, then $\frac{ds}{dt}$ is its velocity and $\frac{dv}{dt}$ its acceleration. Conversely, the anti-derivative of acceleration gives velocity, and the anti-derivative of velocity gives displacement.

Example. A bus leaving Kathmandu has displacement $s(t)=5t^2$ km after $t$ hours. Its velocity is $v=\frac{ds}{dt}=10t$ km/h and acceleration $a=\frac{dv}{dt}=10$ km/h$^2$. Conversely, $\int 10t\,dt=5t^2+C$, where $C$ is fixed by the starting position.

Differential coefficient of $\log_{\sin x}x$. Let $y=\log_{\sin x}x=\dfrac{\ln x}{\ln(\sin x)}$. Then

$\frac{dy}{dx}=\frac{\frac1x\ln(\sin x)-\ln x\cdot\cot x}{\left[\ln(\sin x)\right]^2}=\frac{1}{x\ln(\sin x)}-\frac{\ln x\,\cot x}{\left[\ln(\sin x)\right]^2}.$

22. (b)

The curve is $y=2+\dfrac{x^2}{4}$. It meets $y=11$ where $\dfrac{x^2}{4}=9$, i.e. $x=\pm6$. The region between the $y$-axis, the curve and $y=11$ lies for $x$ from $0$ to $6$:

$A=\int_0^6\left[11-\left(2+\frac{x^2}{4}\right)\right]dx=\int_0^6\left(9-\frac{x^2}{4}\right)dx=\left[9x-\frac{x^3}{12}\right]_0^6=54-18=36.$

$\text{Area}=36\text{ square units}$