Question 12 Marks
Express $|x+1|<0.5$ in neighbourhood and interval form.
View full question & answer→Question 22 Marks
Find the value of $\lim _{h \rightarrow 0} \frac{(x+h)^{5}-x^{5}}{h}$.
View full question & answer→Question 32 Marks
Find the value of $\lim _{x \rightarrow-2} \frac{x^{7}+128}{x+2}$
View full question & answer→Question 42 Marks
Find the value of $\lim _{x \rightarrow 2} \frac{x^{5}-32}{x-2}$.
View full question & answer→Question 52 Marks
Express $0.001$ neighbourhood of $3$ in modulus and interval form.
View full question & answer→Question 62 Marks
Find the value of $\lim _{x \rightarrow 2}\left[\frac{1}{x-2}-\frac{2}{x^{2}-2 x}\right]$.
View full question & answer→Question 72 Marks
Find the value of $\lim _{x \rightarrow 2} \frac{2 x+3}{x-1}$.
View full question & answer→Question 82 Marks
Find the value of $\lim _{x \rightarrow 0} \frac{x^{2}+5}{x^{2}+2}$
View full question & answer→Question 92 Marks
Express $N (5, 0.2)$ in modulus and interval form.
AnswerComparing $N(5,0.2)$ with $N(a, \delta)$, we get $a=5$ and $\delta=0.2$.
Modulus form : $|x-a|<\delta$
Putting $a=5$ and $\delta=0.2$,
$ N(5,0.2)=|x-5|<0.2 $
Interval form : $(a-\delta, a+\delta)$ Putting $a=5$ and $\delta=0.2$,
$ \begin{aligned} N(5,0.2) &=(5-0.2,5+0.2) \\ &=(4.8,5.2) \end{aligned} $
View full question & answer→Question 102 Marks
Explain the meaning of $x \rightarrow a.$
Answer
- If the value of variable $x$ is brought very close to a number $‘a’$ by increasing or decreasing its value then it can be said that $x$ tends to $(A)$
- It is denoted by $x \rightarrow (A)$
- Here $x \rightarrow a$ means, value of $x$ approaches very close to a but it will not be equal to $(A)$
View full question & answer→Question 112 Marks
If $\left|x-A_1\right|<0.09=\left(A_2 ; 4.09\right)$ then find the value of $A_1$ and $A_2$.
AnswerComparing $\left|x-A_1\right|<0.09$ with $|x-a|<\delta$ we get $\mathrm{a}=\mathrm{A}_1$ and $\delta=0.09$
$\mathrm{N}(\mathrm{a} \delta)=(\mathrm{a}-\delta ; \mathrm{a}+\delta)$
$\mathrm{N}\left(\mathrm{~A}_1, 0.09\right)=\left(\mathrm{A}_1-0.09 ; \mathrm{A}_1+0.09\right)$
Hence, $\mathrm{A}_1-0.09=\mathrm{A}_2$ and $\mathrm{A}_1+0.09=4.09$
$\mathrm{A}_1+0.09=4.09$
$\mathrm{~A}_1=4.09-0.09$
$\mathrm{~A}_1=4$
Putting $A_1=4$ in
$\mathrm{A}_1-0.09=\mathrm{A}_2$
$4-0.09=\mathrm{A}_2$
$\mathrm{~A}_2=3.91$
$\mathrm{~A}_2=3.91$
View full question & answer→Question 122 Marks
Express $|3x + 1| < 2$ in neighbourhood and interval form.
Answer$
\begin{aligned}
& |3 x+1|<2 \\
& \therefore\left|x+\frac{1}{3}\right|<\frac{1}{3} \\
& \therefore\left|x-\left(-\frac{1}{3}\right)\right|<\frac{2}{3} \\
& \therefore\left|x-\left(-\frac{1}{3}\right)\right|<\frac{2}{3}, \\
& \therefore a=-\frac{1}{3} \delta=\frac{2}{3}
\end{aligned}
$
In neighbourhood form: $N(a, \delta)$
$\therefore$ Neighbourhood form of $|3 \mathrm{x}+1|<2$
$
=\mathrm{N}\left(-\frac{1}{3}, \frac{2}{3}\right)
$
In Interval form: $(a-\delta, a+\delta)$
Interval form of $13 x+11<2$
$
\begin{aligned}
& =\left(-\frac{1}{3}-\frac{2}{3},-\frac{1}{3}+\frac{2}{3}\right) \\
& =\left(-1, \frac{1}{3}\right)
\end{aligned}
$
View full question & answer→Question 132 Marks
If $N\left(K_1, 0.5\right)=\left(19.5 ; K_2\right)$ then find the value of $K_1$ and $K_2$.
AnswerComparing $N\left(K_1, 0.5\right)=\left(19.5 ; K_2\right)$ with $N(a, \delta)=(a-\delta ; a+\delta)$ we get, $\mathrm{K}_1-0.5=19.5$ and $\mathrm{K}_1+0.5=\mathrm{K}_2$
$\mathrm{K}_1=19.5+0.5$
$\mathrm{K}_1=20$
Now, putting $\mathrm{K}_1=20$ in $\mathrm{K}_1+0.5=\mathrm{K}_2$
$20+0.5=\mathrm{K}_2$
$\mathrm{~K}_2=20.5$
Hence, $K_1=20$ and $K_2=20.5$
View full question & answer→Question 142 Marks
Express the interval form $(-8.75, -7.25)$ in neighbourhood form.
Answer
| Comparing $(-8.75, -7.25)$ with $(a - δ; a + δ)$ we get |
| $a - δ = -8.75$ |
| $a + δ = -7.28$ |
| $2a = -16$ |
| $a = -8$ |
| Putting $a = -8$ in $a + δ = -7.25$ |
| $-8 + δ = -7.25$ |
| $δ = 8 - 7.25$ |
| $δ = 0.75$ |
| Neighbourhoods form: $N(a, δ)$ |
| Putting $a = -8$ and $δ = 0.75$ |
| $(-8.75, -7.25) = N(-8, 0.75)$ |
View full question & answer→Question 152 Marks
Express the interval form $(-0.5, 0.5)$ in modulus form.
Answer
| Comparing $(-0.5, 0.5)$ with $(a - δ ; a + δ)$ we get |
| $a - δ = -0.5$ |
| $a + δ = 0.5$ |
| $2a = 0$ |
| $a = 0$ |
| Putting $a = 0$ in $a + δ = 0.5$ |
| $0 + δ = 0.5$ |
| $δ = 0$ |
| Modulus form: $|x - a| < δ$ |
| Putting $a = 0$ and $δ = 0.5$ |
| $(-0.5, 0.5) = |x| < 0.5$ |
View full question & answer→Question 162 Marks
Define the Punctured $\delta$ neighbourhood of $a.$
Answer
- If $a \in R$ and $δ$ is non-negative real number then the open interval $(a -δ ; a + δ) - \{a\}$ is called punctured δ neighbourhood of $(A)$ It is denoted by $N*(a, δ).$
- $N*(a, δ) = N*(a, δ) - \{a\}$
$= \{x/a -δ < x < a + δ; x \neq a; x \in R\}$
$= \{x/|x-a| < δ; x \neq a; x \in R\}$ View full question & answer→Question 172 Marks
Define the $\delta$ neighborhood of $a.$
Answer
- If $a \in R$ and $δ$ is non-negative real number then the open interval $(a - δ; a + δ)$ is called neighbourhood of $(A)$ It is denoted by $N(a, δ).$
- $N(a, δ) = \{x/a - δ < x < a + δ; x \in R\}$
$= \{x/|x-a| < δ; x \in R\}$ View full question & answer→Question 182 Marks
State the standard form of limit of a polynominal.
AnswerSuppose, $f(x)=a_0+a_1 x+a_2 x^2+\ldots \ldots+a_n x^n$ is a polynomial. The standard form of limit of a polynomial is as follows:
$
\begin{aligned}
\lim _{x \rightarrow b} f(x) & =\lim _{x \rightarrow b}\left(a_0+a_1 x+a_2 x^2+\ldots+a_n x^n\right) \\
& =a_0+a_1 b+a_2 b^2+\ldots+a_n b^n
\end{aligned}
$
View full question & answer→Question 192 Marks
State the division working rule of limit.
Answer$f(x)$ and $g(x)$ are two functions of real variable $x$ and $\lim\limits_{x \rightarrow a} f(x)=1, \lim\limits_{x \rightarrow a} g(x)=m$.
If $\frac{f(x)}{g(x)}$ is the division of the two functions. then division working rule of limit is as follows:
$
\lim\limits_{x \rightarrow a}\left[\frac{f(x)}{g(x)}\right]=\frac{\lim\limits_{x \rightarrow a} f(x)}{\lim\limits_{x \rightarrow a} g(x)}=\frac{l}{m}, \underset{}{\neq 0}
$
Thus, the limit of division of two functions is equal to the division of their limits, where the limit of the function in denominator is not zero.
View full question & answer→Question 202 Marks
State the multiplication working rule of limit.
Answer$f(x)$ and $g(x)$ are two functions of real variable $x$ and $\lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m$.
If $f(x) g \cdot(x)$ is the product of two functions, then multiplication working rule of limit is as follows:
$
\begin{aligned}
\lim _{x \rightarrow a}[f(x) \cdot g(x)] & =\lim _{x \rightarrow a} f(x) \cdot \lim _{x \rightarrow a} g(x) \\
& =l \times m \quad \\
\end{aligned}
$
Thus, limit of product of two functions is equal to the product of limit of these two functions.
View full question & answer→Question 212 Marks
Define limit of a function.
Answer
- The function $f(x)$ has a limit $l$ as $x$ tends to $‘a’$ if for each given predetermined $ε > 0;$ however small, we can find a positive number $δ$ such that $ |f(x) - l | > ε ($Epsilon$)$ for all $x$ such that $|x - a| < δ .$
View full question & answer→Question 222 Marks
Explain the meaning of $x \rightarrow 0.$
Answer
- If by decreasing the positive value of a variable $x$ or by increasing negative value of the variable $x;$ the value of $x$ is brought very close to $‘0’$ then it can be said that $x$ tends to $0.$
- It is denoted by $x \rightarrow 0.$ Here, $x \rightarrow 0$ means the value of $x$ approaches very close to $0$ but it will not be equal to $0.$
View full question & answer→Question 232 Marks
Define open-closed interval.
Answer
- If $a\ R, b\ R$ and $a < b$ then the set of all real numbers between $a$ and $b$ not including $a$ but including $b$ is called an open-closed interval.
- It is denoted by $(a, b]$
- $(a, b] = \{x/a < x \leq b; x \in R\}$
View full question & answer→Question 242 Marks
If $| x -10|$
Answer$|x-10| \therefore\left(10-k_1, 10+k_1\right)=\left(k_2, 10.01\right)$
$\therefore 10+k_1=10.01$
$\therefore k_1=10.01-10=0.01$
Now, putting $k_1=0.01$ in $10-k_1=k_2$
$10-0.01=\mathrm{k}_2$
$\therefore \mathrm{k}_2=9.99$
Hence, $\mathrm{k}_1=0.01$ and $\mathrm{k}_2=9.99$
View full question & answer→Question 252 Marks
If $N (3, b) = (2.95, k),$ then find the values of $b$ and $k.$
Answer$N(3, b)=(2.95, k)$
$ \therefore(a=3, \delta=b)=(a-\delta=2.95, a+\delta=k)$
Putting $a=3$ in $a-\delta=2.95$,
$3-\delta=2.95$
$ \therefore 3-2.95=\delta$
$ \delta=0.05$
But $\delta=b$
$\therefore b=0.05$
Now, $a+\delta=k$
Putting $a=3 ; \delta=0.05$,
$3+0.05=\mathrm{k}$
$ \therefore \mathrm{k}=3.05$
Hence, $b=0.05$ and $k=3.05$
View full question & answer→Question 262 Marks
Express $N(16, 0.5)$ in the interval and modulus form.
Answer$\mathrm{N}(16,0.5) \therefore a=16, \delta=0.5$
In Interval form: $(a-\delta, a+\delta)$
Putting $a=16 ; \delta=0.5$,
$N(16,0.5)=(16-0.5,16+0.5)$ $=(15.5,16.5)$
In Modulus form: $|x-a|<\delta$
Putting $a=16 ; \delta=0.5$,
$N(16,0.5)=|x-16|<0.5$
View full question & answer→Question 272 Marks
Express the following in modulus and neighbourhood form: $1.998 < x < 2.002$
AnswerIn Modulus form: $|x-2|< 0.002$
In Neighbourhood form : $N(2, 0.002)$
View full question & answer→Question 282 Marks
Express the following in modulus and neighbourhood form$: -0.4 < x < 1.4$
AnswerIn Modulus form$:|x-0.5|< 0.9$
In Neighbourhood form $: N(0.5, 0.9)$
View full question & answer→Question 292 Marks
Express the following in modulus and neighbourhood form: $1.95 < x < 2.05$
AnswerIn Modulus form: $|x-2|$
In Neighbourhood form : $N(2, 0.05)$
View full question & answer→Question 302 Marks
Express the following in modulus and neighbourhood form: $3.8 < x < 4.8$
AnswerIn Modulus form $:|x- 4.3|< 0.5$
In Neighbourhood form $:N(4.3, 0.5)$
View full question & answer→Question 312 Marks
Define closed-open interval.
Answer
- If $a \in R, b \in R$ and $a < b$ then the set of all real numbers between $a$ and $b$ including a but not including $b$ is called $a$ closed-open interval.
- It is denoted by $[a, b).$
- $[a, b) = \{x/a \leq x < b; x \in R\}$
View full question & answer→Question 322 Marks
Answer
- If $a \in R, b \in R$ and $a < b$ then the set of all real numbers between $a$ and $b$ including $a$ and $b$ is called a closed interval.
- It is denoted by $[a; b]$
- $[a, b] = \{x/a \leq x \leq b; x \in R\}$
View full question & answer→Question 332 Marks
If $\lim\limits_{x \rightarrow 2} \frac{3}{5 x+K}=\frac{1}{6}$ then find the value of $K$.
Answer
| $\lim _{x \rightarrow 2} \frac{3}{5 x+k}=\frac{1}{6}$ |
| $\therefore \frac{3}{5 x+k}=\frac{1}{6}$ |
| $=\frac{3}{5(2)+K}=\frac{1}{6}($ putting $x=2)$ |
| $=\frac{3}{10+K}=\frac{1}{6}$ |
| $\therefore\ 10+K=18$ |
| $\therefore K=18-10$ |
| $\therefore K=8$ |
View full question & answer→Question 342 Marks
Find the value of $lim_{x\rightarrow-2} \frac{x^{7}+128}{x+2}$
AnswerUsing formula $lim_{x\rightarrow a} \frac{x^n - a^n}{x-a} = n a^{n-1}$
Here $a = -2, n = 7$.
$lim_{x\rightarrow-2} \frac{x^7 - (-2)^7}{x - (-2)} = 7(-2)^{7-1} = 7(-2)^6 = 7 \times 64 = 448$.
View full question & answer→Question 352 Marks
Express $N(16,0.5)$ in the interval and modulus form.
AnswerHere $a = 16$ and $\delta = 0.5$.
Interval Form: $(a-\delta, a+\delta) = (16-0.5, 16+0.5) = (15.5, 16.5)$.
Modulus Form: $|x - a| < \delta \implies |x - 16| < 0.5$.
View full question & answer→Question 362 Marks
Explain the meaning of $x \rightarrow a$.
AnswerThe notation $x \rightarrow a$ (x tends to a) means that the value of variable x is continuously changed such that it comes very close to 'a' from either side, but x never becomes exactly equal to 'a'.
View full question & answer→Question 372 Marks
AnswerLet $a, b \in R$ and $a < b$. The set of all real numbers between a and b, excluding a and b, is called an open interval and is denoted by (a, b). Symbolically, $(a, b) = \{x | a < x < b, x \in R\}$.
View full question & answer→Question 382 Marks
Explain the meaning of $ x\rightarrow0 $.
Answer$ x \rightarrow 0 $ means that as the variable $ x $ takes values closer and closer to 0 from either the left or the right side, the difference between $ x $ and 0 becomes very small, but $ x $ never actually becomes 0.
View full question & answer→Question 392 Marks
If $ N(K_{1}, 0.5) = (19.5, K_{2}) $ then find the value of $ K_{1} $ and $ K_{2} $.
AnswerNeighborhood notation: $ (a - \delta, a + \delta) $.
Here $ a = K_{1} $ and $ \delta = 0.5 $.
$ K_{1} - 0.5 = 19.5 \Rightarrow K_{1} = 20 $.
$ K_{1} + 0.5 = K_{2} \Rightarrow 20 + 0.5 = 20.5 $.
So, $ K_{1} = 20, K_{2} = 20.5 $.
View full question & answer→Question 402 Marks
Express 0.01 neighbourhood of 3 in modulus and interval form.
AnswerModulus form : $|x - 3| < 0.01$
Interval form : $(3 - 0.01, 3 + 0.01) = (2.99, 3.01)$
View full question & answer→