We would like to state that the questions mentioned here have been taken from Basic Mathematics for Grade XI book by SUKUNDA PUSTAK VAWAN, Kathmandu.


1. Find the following limits:

a) $\displaylines{\lim\\^{x \to 2}} (2x^2+3x-14)$

Solution:
Here,
$\displaylines{\lim\\^{x \to 2}} (2x^2+3x-14)$

$= [ 2×(2)^2 + 3×2 - 14]$
$= [ 2×4 + 6 - 14]$
$= [ 8 - 8]$
$= 0$

b) $\displaylines{\lim\\^{x \to 5}}(x^2 + 2x -9)$

Solution:
Here,
$\displaylines{\lim\\^{x \to 5}}(x^2 + 2x -9)$

$= [5^2 + 2×5 - 9]$
$= [25 + 10 - 9]$
$= [35 - 9]$
$= 26$

c) $\displaylines{\lim\\^{x \to 1}} \dfrac{3x^2 + 2x -4}{x^2 + 5x -4}$

Solution:
Here,
$\displaylines{\lim\\^{x \to 1}} \dfrac{3x^2 + 2x -4}{x^2 + 5x -4}$

$= \dfrac{3(1)^2 + 2×1 - 4}{(1)^2 + 5×1 - 4}$

$= \dfrac{3+2-4}{1+5-4}$
$= \dfrac{1}{2}$

d) $\displaylines{ \lim\\^{x \to 3} } \dfrac{6x^2 + 3x -12}{2x^2 + x+1}$

Solution:
Here,
$\displaylines{ \lim\\^{x \to 3} } \dfrac{6x^2 + 3x -12}{2x^2 + x+1}$

$= \dfrac{6(3)^2 + 3(3) -12}{2(3)^2+3+1}$

$= \dfrac{6×9 + 9-12}{2×9 + 4}$

$= \dfrac{54-3}{18+4}$
$= \dfrac{51}{22}$


2. Compute the following limits:

a) $\displaylines{lim\\^{x \to 0}} \dfrac{4x^3-x^2+2x}{3x^2+4x}$
Solution:
Here,
$\displaylines{lim\\^{x \to 0}} \dfrac{4x^3-x^2+2x}{3x^2+4x}$

$= \displaylines{lim\\^{x \to 0}} \dfrac{x(4x^2-x+2)}{x(3x +4)}$

$= \displaylines{lim\\^{x \to 0}}\dfrac{4x^2+x+2}{3x+4}$

$= \dfrac{4(0)^2 - 0 + 2}{3×0 + 4}$

$= \dfrac{2}{4}$
$= \dfrac{1}{2}$

b) $\displaylines{lim\\^{x \to 4} } \dfrac{x^3-64}{x^2-16}$

Solution:
Here,

$\displaylines{lim \\^{x \to 4}}\dfrac{x^3 - 64}{x^2 - 16}$

$= \displaylines{lim\\^{x \to 4} }\dfrac{(x-4)(x^2 + 4x + 16)}{(x-4)(x+4)}$

$= \displaylines{lim\\^{x \to 4}}\dfrac{x^2 + 4x + 16}{x+4}$

$= \dfrac{4^2 + 4*4 + 16}{4+4}$

$= \dfrac{16 + 16 + 16}{8}$

$= \dfrac{16 * 3 }{8}$

$= 2 * 3 $

$= 6$


c) $\displaylines{lim\\^{x \to a}}\dfrac{x⅔-a⅔}{x - a}$

Solution:
Here,

$\displaylines{lim\\^{x \to a}}\dfrac{x⅔-a⅔}{x - a}$

Using formula $na^{n-1}$

$= \dfrac{2}{3} × a^{\frac{2}{3} -1}$

$= \dfrac{2}{3} × a^{\frac{2-3}{3}}$

$= \dfrac{2}{3} a^\frac{-1}{3}$


d) $\displaylines{lim \\ x \to 1} \dfrac{x^2+3x-4}{x-1}$

Solution:
Here,

$\displaylines{lim \\ x \to 1} \dfrac{x^2+3x-4}{x-1}$

The function takes 0/0 form at $x=1$. So,

$= \displaylines{lim \\ x \to 1} \dfrac{(x-1)(x+4)}{(x-1)}$

$= \displaylines{lim \\x \to 1} (x+4)$

$= (1+4)$

$= 5$


e) $\displaylines{lim \\ x \to 2} \dfrac{x^2-5x+6}{x^2-x-2}$

Solution:
Here,

The function takes 0/0 form at $x =2$. So,

$\displaylines{lim \\x \to 2} \dfrac{(x-2)(x-3)}{(x-2)(x+1)}$

$= \displaylines{lim \\ x \to 2} \dfrac{x-3}{x+1}$

$= \dfrac{2-3}{2+1}$

$= - \dfrac{1}{3}$


f) $\displaylines{ lim \\ x \to a} \dfrac{ \sqrt{3x} - \sqrt{2x +a} }{ 2(x -a)}$

Solution:
Here,

The function takes 0/0 form at $x = a$. So,

$= \displaylines{ lim \\ x \to a} \dfrac{ \sqrt{3x} - \sqrt{2x + a} }{2(x -a)} × \dfrac{ \sqrt{3x} + \sqrt{2x +a} }{ \sqrt{3x} + \sqrt{2x +a} }$

$= \displaylines{ lim \\ x \to a} \dfrac{3x - (2x + a)}{2(x-a) ( \sqrt{3x} + \sqrt{2x +a} )}$

$= \displaylines{ lim \\ x \to a} \dfrac{3x - 2x - a}{2(x-a) (\sqrt{3x} + \sqrt{2x + a}) }$

$= \displaylines{ lim \\ x \to a} \dfrac{(x-a)}{2(x-a) (\sqrt{3x} + \sqrt{2x + a})}$

$= \displaylines{ lim \\ x \to a} \dfrac{1}{ 2(\sqrt{3x} + \sqrt{2x + a} ) }$

$= \dfrac{1}{2(\sqrt{3×a} + \sqrt{2×a + a} )}$

$= \dfrac{1}{2(\sqrt{3a} + \sqrt{3a})}$

$= \dfrac{1}{2×2 \sqrt{3a}}$

$= \dfrac{1}{4 \sqrt{3a}}$



3. Calculate the following limits:

a) $\displaylines{\lim\\^{x \to \infty}}\dfrac{2x^2}{3x^2+2}$


Solution:

$\displaylines{\lim\\^{x \to \infty}}\dfrac{2x^2}{3x^2+2}$

$= \displaylines{\lim \\^{x \to \infty}}\dfrac{x^2(2)}{x^2(3 + \frac{2}{x^2})}$

$= \displaylines{\lim \\^{x \to \infty}} \dfrac{2}{3+ \frac{2}{x^2}}$

$= \dfrac{2}{3 + \frac{2}{\infty^2}}$

$= \dfrac{2}{3+ 0}$

$= \dfrac{2}{3}$



More about the publishers of the book:
SUKUNDA PUSTAK BHAWAN
Bhotahity, Kathmandu
5320379, 5353537