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{xâ…”-aâ…”}{x - a}

Solution:
Here,

\displaylines{lim\\^{x \to a}}\dfrac{xâ…”-aâ…”}{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