Download App

Statistics 1 Marks Each

LIMIT · Gujarat Board (GSEB) STD 12 Commerce (GSEB - English Medium). 72 questions.

Questions · Page 2 of 2

1 Marks Each

Question 511 Mark
Find the value of $\lim _{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x}$.
Answer
$\lim _{x \rightarrow 1} \frac{1-\sqrt{x}}{1-x}$
$=\lim _{x \rightarrow 1} \frac{1-\sqrt{x}}{(1-\sqrt{x})(1+\sqrt{x})}$
$=\lim _{x \rightarrow 1} \frac{1}{1+\sqrt{x}}$
$=\frac{1}{1+\sqrt{1}}$
$=\frac{1}{1+1}$
$=\frac{1}{2}$
View full question & answer→
Question 521 Mark
Find the value of $\operatorname{Lim}_{x \rightarrow 0} \frac{3 x^2-4 x+10}{2 x+5}$
Answer
$\operatorname{Lim}_{x \rightarrow 0} \frac{3 x^2-4 x+10}{2 x+5}$
$\therefore \frac{3(0)^2-4(0)+10}{2(0)+5}$
$\therefore \frac{0+0+10}{0+5}$
$\therefore\frac{10}{5}$
View full question & answer→
Question 531 Mark
Find the value of $\lim _{x \rightarrow 1} \frac{x-1}{x^2-1}$.
Answer
$\lim _{x \rightarrow 1} \frac{x-1}{x^2+1}$
$=\frac{1-1}{(1)^2-1}$
$=\frac{0}{1+1}$
$=\frac{0}{2}=0$
View full question & answer→
Question 541 Mark
If $\lim _{x \rightarrow-2} 3 x+k=8$ then find the value of $k$.
Answer
$\lim _{x \rightarrow-2} 3x + K = 83$
$3x + K = 8$
$3(-2) + K = 8$
$-6 + K = 8$
$K = 8 + 6$
$K = 14$
View full question & answer→
Question 551 Mark
Find the value of $\lim _{x \rightarrow 5} \frac{x^3-125}{x-5}$.
Answer
$\lim _{x \rightarrow 5} \frac{x^3-125}{x-5}$
$\lim _{\mathrm{x} \rightarrow 5} \frac{\mathrm{x}^3-5^3}{x-5}\left[\lim _{x \rightarrow \mathrm{a}} \frac{\mathrm{x}^n-5^{\mathrm{n}}}{\mathrm{x}-\mathrm{a}}=n a^{\mathrm{n}-1}\right]$
$3(5)^{3-1}$
$3(5)^2$
$3 \times 25=75$
View full question & answer→
Question 561 Mark
Find the value of $\lim _{x \rightarrow 0} \frac{2 x^2-3 x+8}{3 x+4}$.
Answer
$\lim _{x \rightarrow 0} \frac{2 x^2-3 x+8}{3 x+4}$
$\lim _{x \rightarrow 0} \frac{2(0)^2-3(0)+8}{3(0)+4}$
$\frac{0+0+8}{0+4}$
$\frac{8}{4}$
$=2$
View full question & answer→
Question 571 Mark
Find the value of $\operatorname{Lim}_{x \rightarrow-2} \sqrt[2]{10-3 x}$.
Answer
$\lim _{x \rightarrow 2} \sqrt[2]{10-3 x}$
$\sqrt[2]{10-3 x(-2)}$
$\sqrt[2]{10+6}$
$\sqrt[2]{16}$
$= 4$
View full question & answer→
Question 581 Mark
If $|x + 3| < 0.05 = (a, -2.95)$ then find the value of $a.$
Answer
Comparing $|x + 3| < 0.05$ with $|x - a| < δ;$ we get $a = -3$ and $δ = 0.05$
Now, $N(a, δ) = (a - δ; a + δ)$
$N(-3; 0.05) = (-3 0.05; -3 + 0.05)$
$=(-3.05 ;-2.95)$
Hence, $(a, -2.95) = (-3.05; -2.95)$
$\therefore a = -3.05$
View full question & answer→
Question 591 Mark
If $N (a; 0.3) = |x - 5|$
Answer
Comparing $N(a; 0.3) = |x - 5| < b$ with $N(a, δ) = |x - a| < δ;$
we get $a = 5$ and $δ =0.3.$
View full question & answer→
Question 601 Mark
Express $|3 x|<\frac{1}{3}$ in interval form.
Answer
Comparing $|3x| < \frac{1}{3}$ with $|x - a|<δ$ we get
$a = 0$ and $δ = \frac{1}{9}$.
Interval form $: N(a - δ ; a + δ)$
Putting $a = 0$ and $δ =\frac{1}{9}$.
$|3 x|<\frac{1}{3}=\left(0-\frac{1}{9} ; 0+\frac{1}{9}\right)$
$=\left(-\frac{1}{9} ; \frac{1}{9}\right)$
View full question & answer→
Question 611 Mark
Express $|x+5|<\frac{1}{2}$ in neighborhood form.
Answer
Comparing $|x+5|< \frac{1}{2}$ with $|x - a|<δ,$ we get
$a = -5$ and $δ = \frac{1}{2}$
Neighbourhood form $: N(a; δ)$
Putting $a = -5$ and $= δ =\frac{1}{2}$
$|x+5|<\frac{1}{2}=N\left(-5 ; \frac{1}{2}\right)$
View full question & answer→
Question 621 Mark
What is the limit of the product of a function with a constant?
Answer
The limit of the product of a function with a constant is equal to the product of the limit of the function with the same constant.
That means $\stackrel{\lim }{\mathrm{x} \rightarrow \mathrm{a}} K.$
$f(x) = K.L;$ where $K$ is the constant.
View full question & answer→
Question 631 Mark
Define neighborhood of a.
Answer
Any open interval containing $a, a ∈ R$ is called a neighbourhood of $(A).$
View full question & answer→
Question 651 Mark
Define an interval.
Answer
The real line or real number line is a line Where its points are the real numbers.
View full question & answer→
Question 661 Mark
Define the real line.
Answer
The real line or real number line is a line Where its points are the real numbers.
View full question & answer→
Question 671 Mark
If the value of the function is indeterminate then how can be we obtained the approximate value of the function?
Answer
If the value of the function is indeterminate limit can be used to approximate the value of a function.
View full question & answer→
Question 681 Mark
If $\lim\limits_{x \rightarrow 5} \ 4 x+k=6$ then find the value of $K$.
Answer
$\operatorname{Lim}_{x \rightarrow 5} 4x + k =6$
$\therefore 4x + k = 6 ($putting $x = -1)$
$\therefore 4(-1) + k =6$
$\therefore -4 + k = 6$
$\therefore K + 6 + 4$
$\therefore k = 10$
View full question & answer→
Question 691 Mark
Express $N (-20; 0.4)$ in modulus form.
Answer
Comparing $N(-20; 0.4)$ with $N(a, δ)$
we get $a = -20$ and $δ = 0.4.$
Modulus form $: |x - a| < δ$
Putting $a = -20$ and $δ = 0.4$
$N(-20; 0.4) = |x + 20| < 0.4$
View full question & answer→
Question 701 Mark
When it should be said that limit of a function does not exist?
Answer
When the value of $x$ is brought nearer to a by increasing or decreasing its value, the value of $f(x)$ does not approach to a particular value then it should be said that limit of a function does not exist.
View full question & answer→
Question 711 Mark
If $ lim_{x\rightarrow-1} 4x+k=6 $ then find the value of k.
Answer
Substitute $ x = -1 $ in the equation:
$ 4(-1) + k = 6 $
$ -4 + k = 6 $
$ k = 6 + 4 = 10 $
View full question & answer→
Generate Paper FreeAll LIMIT topics
1 Marks Each - Page 2 - Statistics STD 12 Commerce Questions - Vidyadip