Limits and Derivatives
Exercise 12.1 Solutions
Standard Limits:
1. $\lim_{x \to a} \frac{x^n – a^n}{x – a} = n a^{n-1}$
2. $\lim_{x \to 0} \frac{\sin x}{x} = 1$
3. $\lim_{x \to 0} \frac{1 – \cos x}{x} = 0$
1. $\lim_{x \to a} \frac{x^n – a^n}{x – a} = n a^{n-1}$
2. $\lim_{x \to 0} \frac{\sin x}{x} = 1$
3. $\lim_{x \to 0} \frac{1 – \cos x}{x} = 0$
Algebraic Limits
Evaluate the following limits:
6. $\lim_{x \to 0} \frac{(x+1)^5 – 1}{x}$
7. $\lim_{x \to 2} \frac{3x^2 – x – 10}{x^2 – 4}$
8. $\lim_{x \to 3} \frac{x^4 – 81}{2x^2 – 5x – 3}$
10. $\lim_{z \to 1} \frac{z^{1/3} – 1}{z^{1/6} – 1}$
1. $x \to 3$
$\lim_{x \to 3} (x + 3) = 3 + 3 = 6$
2. $x \to \pi$
$\lim_{x \to \pi} (x – \frac{22}{7}) = \pi – \frac{22}{7}$
3. $r \to 1$
$\lim_{r \to 1} \pi r^2 = \pi(1)^2 = \pi$
4. $x \to 4$
$\lim_{x \to 4} \frac{4x+3}{x-2} = \frac{4(4)+3}{4-2} = \frac{19}{2}$
5. $x \to -1$
$\lim_{x \to -1} \frac{x^{10}+x^5+1}{x-1} = \frac{(-1)^{10}+(-1)^5+1}{-1-1} = \frac{1-1+1}{-2} = -\frac{1}{2}$
6. $\lim_{x \to 0} \frac{(x+1)^5 – 1}{x}$
Sub $y=x+1$
As $x \to 0, y \to 1$. Limit becomes $\lim_{y \to 1} \frac{y^5 – 1^5}{y-1}$.
Formula
Using $\lim_{y \to a} \frac{y^n-a^n}{y-a} = na^{n-1}$. Here $n=5, a=1$.
Result
$5(1)^{5-1} = 5$.
7. $\lim_{x \to 2} \frac{3x^2 – x – 10}{x^2 – 4}$
Factorize
Num: $3x^2 – 6x + 5x – 10 = 3x(x-2)+5(x-2) = (3x+5)(x-2)$
Denom: $(x-2)(x+2)$
Denom: $(x-2)(x+2)$
Limit
$\lim_{x \to 2} \frac{3x+5}{x+2} = \frac{3(2)+5}{2+2} = \frac{11}{4}$
8. $\lim_{x \to 3} \frac{x^4 – 81}{2x^2 – 5x – 3}$
Factorize
Num: $(x^2-9)(x^2+9) = (x-3)(x+3)(x^2+9)$
Denom: $2x^2-6x+x-3 = (2x+1)(x-3)$
Denom: $2x^2-6x+x-3 = (2x+1)(x-3)$
Limit
$\lim_{x \to 3} \frac{(x+3)(x^2+9)}{2x+1} = \frac{(6)(18)}{7} = \frac{108}{7}$
10. $\lim_{z \to 1} \frac{z^{1/3} – 1}{z^{1/6} – 1}$
Structure
Divide Num and Denom by $(z-1)$. $\frac{\frac{z^{1/3}-1}{z-1}}{\frac{z^{1/6}-1}{z-1}}$
Apply Formula
$\frac{\frac{1}{3}(1)^{-2/3}}{\frac{1}{6}(1)^{-5/6}} = \frac{1/3}{1/6} = \frac{6}{3} = 2$.
Trigonometric & Mixed
11. $\lim_{x \to 1} \frac{ax^2+bx+c}{cx^2+bx+a}$ ($a+b+c \ne 0$)
15. $\lim_{x \to \pi} \frac{\sin(\pi – x)}{\pi(\pi – x)}$
17. $\lim_{x \to 0} \frac{\cos 2x – 1}{\cos x – 1}$
18. $\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}$
20. $\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}$
21. $\lim_{x \to 0} (\text{cosec } x – \cot x)$
22. $\lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x – \frac{\pi}{2}}$
Direct Sub
$\frac{a(1)^2+b(1)+c}{c(1)^2+b(1)+a} = \frac{a+b+c}{a+b+c} = 1$.
12. $\lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x+2}$
Simplify
Num: $\frac{2+x}{2x}$. Expr: $\frac{x+2}{2x(x+2)} = \frac{1}{2x}$.
Result
$\frac{1}{2(-2)} = -\frac{1}{4}$.
13. $\lim_{x \to 0} \frac{\sin ax}{bx}$
Adjust
$\frac{\sin ax}{ax} \times \frac{ax}{bx}$.
Result
$1 \times \frac{a}{b} = \frac{a}{b}$.
15. $\lim_{x \to \pi} \frac{\sin(\pi – x)}{\pi(\pi – x)}$
Sub $y=\pi-x$
As $x \to \pi, y \to 0$. Limit becomes $\lim_{y \to 0} \frac{\sin y}{\pi y}$.
Result
$\frac{1}{\pi} \lim_{y \to 0} \frac{\sin y}{y} = \frac{1}{\pi}(1) = \frac{1}{\pi}$.
17. $\lim_{x \to 0} \frac{\cos 2x – 1}{\cos x – 1}$
Identity
Use $1 – \cos \theta = 2 \sin^2(\theta/2)$.
Expr becomes $\frac{-(1-\cos 2x)}{-(1-\cos x)} = \frac{2\sin^2 x}{2\sin^2(x/2)}$.
Expr becomes $\frac{-(1-\cos 2x)}{-(1-\cos x)} = \frac{2\sin^2 x}{2\sin^2(x/2)}$.
Expand
$\sin x = 2\sin(x/2)\cos(x/2)$. Square it: $4\sin^2(x/2)\cos^2(x/2)$.
Simplify
$\frac{4\sin^2(x/2)\cos^2(x/2)}{\sin^2(x/2)} = 4\cos^2(x/2)$.
Limit
As $x \to 0, 4\cos^2(0) = 4(1) = 4$.
18. $\lim_{x \to 0} \frac{ax + x \cos x}{b \sin x}$
Divide by x
Num: $a + \cos x$. Denom: $b(\frac{\sin x}{x})$.
Limit
$\frac{a + \cos 0}{b(1)} = \frac{a+1}{b}$.
20. $\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx}$
Divide by x
Num: $\frac{\sin ax}{x} + b \Rightarrow a(\frac{\sin ax}{ax}) + b \to a+b$.
Denom
$a + \frac{\sin bx}{x} \Rightarrow a + b(\frac{\sin bx}{bx}) \to a+b$.
Result
$\frac{a+b}{a+b} = 1$.
21. $\lim_{x \to 0} (\text{cosec } x – \cot x)$
Simplify
$\frac{1}{\sin x} – \frac{\cos x}{\sin x} = \frac{1 – \cos x}{\sin x}$.
Half Angle
$\frac{2\sin^2(x/2)}{2\sin(x/2)\cos(x/2)} = \frac{\sin(x/2)}{\cos(x/2)} = \tan(x/2)$.
Limit
$\tan(0) = 0$.
22. $\lim_{x \to \frac{\pi}{2}} \frac{\tan 2x}{x – \frac{\pi}{2}}$
Sub $h=x-\frac{\pi}{2}$
As $x \to \frac{\pi}{2}, h \to 0$. Also $x = \frac{\pi}{2} + h$.
Num
$\tan 2(\frac{\pi}{2} + h) = \tan(\pi + 2h) = \tan 2h$.
Limit
$\lim_{h \to 0} \frac{\tan 2h}{h} = \lim_{h \to 0} \frac{\tan 2h}{2h} \times 2 = 1 \times 2 = 2$.
Limits and Derivatives
Exercise 12.1 Solutions (Q23 – Q32)
Q23
23. Find $\lim_{x \to 0} f(x)$ and $\lim_{x \to 1} f(x)$, where $f(x) = \begin{cases} 2x+3, & x \le 0 \\ 3(x+1), & x > 0 \end{cases}$
(i) Limit at $x \to 0$
L.H.L
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0} (2x + 3) = 2(0) + 3 = 3$
R.H.L
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0} 3(x+1) = 3(0+1) = 3$
Result
Since L.H.L = R.H.L, $\lim_{x \to 0} f(x) = 3$.
(ii) Limit at $x \to 1$
Check Domain
For $x \to 1$, $x > 0$. We use $f(x) = 3(x+1)$.
Limit
$\lim_{x \to 1} 3(x+1) = 3(1+1) = 6$.
Limit at 0 is 3; Limit at 1 is 6
Q24
24. Find $\lim_{x \to 1} f(x)$, where $f(x) = \begin{cases} x^2-1, & x \le 1 \\ -x^2-1, & x > 1 \end{cases}$
L.H.L
$\lim_{x \to 1^-} (x^2 – 1) = 1^2 – 1 = 0$
R.H.L
$\lim_{x \to 1^+} (-x^2 – 1) = -(1)^2 – 1 = -2$
Conclusion
L.H.L $\ne$ R.H.L ($0 \ne -2$).
Limit does not exist
Q25
25. Evaluate $\lim_{x \to 0} f(x)$, where $f(x) = \begin{cases} \frac{|x|}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$
L.H.L ($x<0$)
$|x| = -x$. Limit is $\lim_{x \to 0} \frac{-x}{x} = -1$.
R.H.L ($x>0$)
$|x| = x$. Limit is $\lim_{x \to 0} \frac{x}{x} = 1$.
Result
$-1 \ne 1$. Limit does not exist.
Limit does not exist
Q26
26. Find $\lim_{x \to 0} f(x)$, where $f(x) = \begin{cases} \frac{x}{|x|}, & x \ne 0 \\ 0, & x = 0 \end{cases}$
L.H.L
$\lim_{x \to 0^-} \frac{x}{-x} = -1$.
R.H.L
$\lim_{x \to 0^+} \frac{x}{x} = 1$.
Limit does not exist
Q27
27. Find $\lim_{x \to 5} f(x)$, where $f(x) = |x| – 5$
Substitute
At $x=5$, $|x| = 5$.
Result
$|5| – 5 = 5 – 5 = 0$.
0
Q28
28. Find $a, b$ if $\lim_{x \to 1} f(x) = f(1)$, where $f(x) = \begin{cases} a+bx, & x<1 \\ 4, & x=1 \\ b-ax, & x>1 \end{cases}$
L.H.L
$\lim_{x \to 1^-} (a+bx) = a+b$.
R.H.L
$\lim_{x \to 1^+} (b-ax) = b-a$.
Condition
Limit exists and equals $f(1)=4$. So, $a+b=4$ and $b-a=4$.
Solve
Adding equations: $2b=8 \Rightarrow b=4$. Substituting: $a+4=4 \Rightarrow a=0$.
$a = 0, \quad b = 4$
Q29
29. Let $f(x) = (x-a_1)(x-a_2)\dots(x-a_n)$. Find $\lim_{x \to a_1} f(x)$ and $\lim_{x \to a} f(x)$.
At $x \to a_1$
$\lim_{x \to a_1} (x-a_1)(x-a_2)\dots = (a_1-a_1)(a_1-a_2)\dots = 0 \times \dots = 0$.
At $x \to a$
$\lim_{x \to a} f(x) = (a-a_1)(a-a_2)\dots(a-a_n)$.
Limit at $a_1$ is 0; Limit at a is $\prod (a-a_i)$
Q30
30. If $f(x) = \begin{cases} |x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>0 \end{cases}$. For what ‘a’ does $\lim_{x \to a} f(x)$ exist?
Check 0
LHL: $\lim_{x \to 0} (-x+1) = 1$. RHL: $\lim_{x \to 0} (x-1) = -1$. Limit DNE at $a=0$.
Other ‘a’
For any $a \ne 0$, the function is continuous (either linear $|x|+1$ or $|x|-1$).
Exists for all $a \in \mathbb{R} – \{0\}$
Q31
31. If $\lim_{x \to 1} \frac{f(x)-2}{x^2-1} = \pi$, evaluate $\lim_{x \to 1} f(x)$.
Logic
Since the denominator $x^2-1 \to 0$ as $x \to 1$, and the limit is finite ($\pi$), the numerator must also approach 0.
Condition
$\lim_{x \to 1} (f(x) – 2) = 0$.
Solve
$\lim_{x \to 1} f(x) = 2$.
2
Q32
32. For what integers m, n do $\lim_{x \to 0} f(x)$ and $\lim_{x \to 1} f(x)$ exist?
$f(x) = \begin{cases} mx^2+n, & x<0 \\ nx+m, & 0 \le x \le 1 \\ nx^3+m, & x>1 \end{cases}$
$f(x) = \begin{cases} mx^2+n, & x<0 \\ nx+m, & 0 \le x \le 1 \\ nx^3+m, & x>1 \end{cases}$
At $x=0$
LHL: $m(0)^2+n = n$. RHL: $n(0)+m = m$.
Limit exists if $n = m$.
Limit exists if $n = m$.
At $x=1$
LHL: $n(1)+m = n+m$. RHL: $n(1)^3+m = n+m$.
Limit exists for any integer values of m and n.
Limit exists for any integer values of m and n.
Combine
For limit at 0 to exist, $n=m$. For limit at 1 to exist, any integers work.
Both exist if $m = n$