KC Sinha: Exercise 24.1- Mathematics Solution Class 11 Chapter 24 सीमा

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Question 2

निकाले 

(Evaluate)

$\lim_{x\rightarrow 1}\frac{x^3-1}{x-1}$

Sol :

$=\lim_{x\rightarrow 1}\frac{x^3-1^3}{x-1}$

=3.(1)^3-1

=3(1)²

=3

Question 3

निकाले 

(Evaluate)

$\lim_{x\rightarrow a}\frac{x^4-a^4}{x-a}$
Sol :
$\lim_{x\rightarrow a}\frac{x^n-a^n}{x-a}=n.a^{n-1}$
=4.a^4-1
=4a³

Question 5

निकाले 

(Find)

$\lim_{x\rightarrow -3}\frac{x^3+27}{x^5+243}$
Sol :

$\lim_{x\rightarrow -3}\frac{x^3-(-3)^3}{x^5-(-3)^5}$

अंश तथा हर मे x-(3) से भाग देने पर

$\lim_{x\rightarrow -3}\dfrac{\frac{x^3-(-3)^3}{x-(-3)}} {\frac{x^5-(-3)^5}{x-(-3)}}$

$=\dfrac{\lim_{x\rightarrow -3}\frac{x^3-(-3)^3}{\lim_{x\rightarrow -3}}}{\frac{x^5-(-3)^5}{x-(-3)}}$

$=\frac{3(-3)^{3-1}}{5(-3)^{5-1}}$

$=\frac{3(-3)^2}{5(-3)^4}$

$=\frac{3\times 9}{5\times 81}=\frac{1}{15}$

Question 6

निकाले 

(Find)

(i) $\lim_{r\rightarrow 1}\frac{\sqrt{t}-1}{\sqrt[3]{t}-1}$
Sol :

$=\lim_{t\rightarrow 1}\frac{t^{\frac{1}{2}-1^{\frac{1}{2}}}}{t^{\frac{1}{3}}-1^{\frac{1}{3}}}$

अंश तथा हर मे t-1 से भाग देने पर

$=\lim_{t\rightarrow 1} \dfrac{\frac{t^{\frac{1}{2}}-1^{\frac{1}{2}}}{t-1}}{\frac{t^{\frac{1}{3}}-1^{\frac{1}{3}}}{t-1}}$

$=\dfrac{\lim_{t\rightarrow 1} \frac{t^{\frac{1}{2}}-1^{\frac{1}{2}}}{t-1}}{\lim_{t\rightarrow 1} \frac{t^{\frac{1}{3}}-1^{\frac{1}{3}}}{t-1}}$

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

(ii) $\lim_{x\rightarrow 1}\frac{x^{15}-1}{x^{10}-1}$

Sol :

(iii) $\lim_{x\rightarrow 1}\frac{x^{1/3}-1}{x^{1/6}-1}$

Sol :

$\lim_{x\rightarrow a}\frac{x-a}{x^{3/2}-a^{3/2}}$

Sol :

$=\lim_{x\rightarrow a}\dfrac{1}{\frac{x^{3/2}-a^{3/2}}{x-a}}$

$=\dfrac{1}{\lim_{x\rightarrow a}\frac{x^{3/2}-a^{3/2}}{x-a}}$

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

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

$=\frac{2}{3\sqrt{a}}$

Question 9

$\lim_{x\rightarrow 2}\frac{\sqrt[3]{x-1}-1}{x-2}$

Sol :

$=\lim_{x\rightarrow 2}\frac{(x-1)^{\frac{1}{3}}-1^{\frac{1}{3}}}{(x-1)-1}$

$=\frac{1}{3}(1)^{\frac{1}{3}-1}$

$=\frac{1}{3}$

Question 10

$\lim_{x\rightarrow 0}\frac{(1+x)^n-1}{x}$

Sol :

$=\lim_{x\rightarrow 0}\frac{(1+x)^n-1^n}{(1+x)-1}$

=n(1)^n-1

=n

Question 10

$\lim_{x\rightarrow 0}\frac{(1+x)^{1/3}-(1-x)^{1/3}}{x}$

Sol :

$=\lim_{x\rightarrow 0}\frac{(1+x)^{1/3}-(1-x)^{1/3}}{2x}\times 2$

$=2\lim_{x\rightarrow 0}\frac{(1+x)^{1/3}-(1-x)^{1/3}}{(1+x)-(1-x)}$

$=2\times \frac{1}{3}(1-0)^{\frac{1}{3}-1}$

$=\frac{2}{3}$

Question 14

$\lim_{x\rightarrow 1}\left(\frac{1-x^{-1/3}}{1-x^{-2/3}}\right)$

Sol :

$=\lim_{x\rightarrow 1}\left(\frac{-(x^{\frac{1}{3}}-1)}{-(x^{-\frac{2}{3}}-1)}\right)$

$=\lim_{x\rightarrow 1}\frac{x^{\frac{-1}{3}}-1^{\frac{-1}{3}}}{x^{\frac{-2}{3}}-1^{\frac{-1}{3}}}$

अंश तथा हर मे x-1 से भाग देने पर

$=\lim_{x\rightarrow 1}\dfrac{\frac{x^{-\frac{1}{3}}-1^{-\frac{1}{3}}}{x-1}}{\frac{x^{-\frac{2}{3}}-1^{\frac{-2}{3}}}{x-1}}$

$=\dfrac{\lim_{x\rightarrow 1}\frac{x^{-\frac{1}{3}}-1^{-\frac{1}{3}}}{x-1}}{\lim_{x\rightarrow 1}\frac{x^{-\frac{2}{3}}-1^{\frac{-2}{3}}}{x-1}}$

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

$=\frac{1}{2}$

Question 16

निकाले (Find)

$\lim_{x\rightarrow a}\frac{(x+2)^{5/2}-(a+2)^{5/2}}{x-a}$

Sol :

$=\lim_{x\rightarrow a}\frac{(x+2)^{\frac{5}{2}}-(a+2)^{\frac{5}{2}}}{(x+2)-(a+2)}$

$=\frac{5}{2}(a+2)^{\frac{5}{2}-1}$

$=\frac{5}{2}(a+2)^{3/2}$

Question 17

धनात्मंक पूर्णांक का मान निकाले ताकि

(Find the positive integer n so that )

$\lim_{x\rightarrow 3}\frac{x^n-3^n}{x-3}=108$

Sol :

n(3)^n-1=108

n(3)^n-1=2²×3³

n=4, 

n-1=3

n=4

Question 19

a के सभी संभव मानो को निकाले यदि

(Find all possible values of a, if)

$\lim_{x\rightarrow a}\frac{x^3-a^3}{x-a}=\lim_{x\rightarrow 1}\frac{x^4-1^4}{x-1}$

Sol :

3.a^3-1=4.1^4-1

3a²=4

a²=$\frac{4}{3}$

$a=\pm \sqrt{\frac{4}{3}}$

$a=\pm \frac{2}{\sqrt{3}}$

Question 20

$\lim_{x\rightarrow 0}\frac{\sqrt{1+4x}-\sqrt{1-3x}}{x}$

Sol :

अंश तथा हर मे x-1 से भाग देने पर

$=\lim_{x\rightarrow 0}\frac{\sqrt{1+4x}-\sqrt{1-3x}}{x}\times \frac{\sqrt{1+4x}+\sqrt{1-3x}}{\sqrt{1+4x}+\sqrt{1-3x}}$

$=\lim_{x\rightarrow 0}\frac{(\sqrt{1+4x})^2-(\sqrt{1-3x})^2}{x(\sqrt{1+4x})+\sqrt{1-3x}}$

$=\lim_{x\rightarrow 0}\frac{1+4x-1+3x}{x(\sqrt{1+4x}+\sqrt{1-3x})}$

$=\lim_{x\rightarrow 0}\frac{7x}{x(\sqrt{1+4x}+\sqrt{1-3x})}$

$=\frac{7}{\sqrt{1+4(0)}+\sqrt{1-3(0)}}$

$=\frac{7}{1+1}=\frac{7}{2}$

Question 21

निकाले (Find)

$\lim_{x\rightarrow 0}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x}$

Sol :

अंश का परिमेय करण करने पर,

$=\lim_{x\rightarrow 0}\frac{\sqrt{1+x^2}-\sqrt{1-x^2}}{x}\times \frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}+\sqrt{1-x^2}}$

$=\lim_{x\rightarrow }\frac{(\sqrt{1+x^2})^2-(\sqrt{1-x^2})^2}{x(\sqrt{1+x^2}+\sqrt{1-x^2})}$

$=\lim_{x\rightarrow 0}\frac{1+x^2-1+x^2}{x(\sqrt{1+x^2}+\sqrt{1-x^2})}$

$=\lim_{x\rightarrow 0}\frac{2x^2}{x(\sqrt{1+x^2}+\sqrt{1-x^2})}$

$=\frac{2\times 0}{\sqrt{1+0^2}+\sqrt{1-0^2}}$

$=\frac{0}{2}=0$

Question 22

निकाले (Find)

$\lim_{x\rightarrow 1}\frac{\sqrt{2-x}-1}{1-x}$

Sol :

अंश का परिमेय करण करने पर,

$=\lim_{x\rightarrow 1}\frac{\sqrt{2-x}-1}{1-x}\times \frac{\sqrt{2-x}+1}{\sqrt{2-x}+1}$

$=\lim_{x\rightarrow 1}\frac{(\sqrt{2-x})^2-1^2}{(1-x)(\sqrt{2-x}+1)}$

$=\lim_{x\rightarrow 1}\frac{2-x-1}{(1-x)(\sqrt{2-x}+1)}$

$=\lim_{x\rightarrow 1}\frac{1-x}{(1-x)(\sqrt{2-x}+1)}$

$=\frac{1}{\sqrt{2-1}+1}=\frac{1}{1+1}$

$=\frac{1}{2}$

Question 23

निकाले (Find)

$\lim_{x\rightarrow 0}\frac{\sqrt{1+x}-1}{x}$

Sol :

अंश का परिमेय करण करने पर,

$=\lim_{x\rightarrow 0}\frac{\sqrt{1+x}-1}{x}\times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$

$=\lim_{x\rightarrow 0}\frac{(\sqrt{1+x})^2-1^2}{x(\sqrt{1+x}+1)}$

$=\lim_{x\rightarrow 0}\frac{1+x-1}{x(\sqrt{1+x}+1)}$

$=\lim_{x\rightarrow 0}\frac{x}{x(\sqrt{1+x}+1)}$

$=\frac{1}{\sqrt{1+0}+1}=\frac{1}{2}$

Question 27

निकाले (Find)

$\lim_{h\rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$,x≠0

Sol :

अंश का परिमेय करण करने पर,

$=\lim_{h\rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$

$=\lim_{h\rightarrow 0}\frac{(\sqrt{x+h})^2-(\sqrt{x})^2}{h(\sqrt{x+h}+\sqrt{x})}$

$=\lim_{h\rightarrow 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}$

$=\lim_{h\rightarrow 0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$

$=\frac{1}{\sqrt{x+0}+\sqrt{x}}=\frac{1}{2\sqrt{x}}$

Question 28

निकाले (Find)

$\lim_{x\rightarrow 0}\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a(a+x)}}$

Sol :

अंश का परिमेय करण करने पर,

$=\lim_{x\rightarrow 0}\frac{\sqrt{a+x}-\sqrt{a}}{x\sqrt{a(a+x)}}\times \frac{\sqrt{a+x}+\sqrt{x}}{\sqrt{a+x}+\sqrt{x}}$

$=\lim_{x\rightarrow 0}\frac{(\sqrt{a+x})^2-(\sqrt{a})^2}{x\sqrt{a(a+x)}+\sqrt{a}}$

$=\lim_{x\rightarrow 0}\frac{a+x-a}{x\sqrt{a(a+x)}(\sqrt{a+x}+\sqrt{a})}$

$=\lim_{x\rightarrow 0}\frac{x}{x\sqrt{a(a+x)}(\sqrt{a+x}+\sqrt{a})}$

$=\frac{1}{\sqrt{a(a+0)}(\sqrt{a+0}+\sqrt{a})}$

$=\frac{1}{a(2\sqrt{a})}=\frac{1}{2a\sqrt{a}}$

$=\frac{1}{2a^1.a^{\frac{1}{2}}}$

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

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

Question 29

निकाले (Find)

$\lim_{x\rightarrow 0}\frac{(2x-3)(\sqrt{x}-1)}{2x^2+x-3}$

Sol :

$=\lim_{x\rightarrow 0}\frac{(2x-3)(\sqrt{x}-1)}{2x^2+3x-2x-3}$

$=\lim_{x\rightarrow 0}\frac{(2x-3)(\sqrt{x}-1)}{x(2x+3)-1(2x+3)}$

$=\lim_{x\rightarrow 0}\frac{(2x-3)(\sqrt{x}-1)}{(2x+3)(x-1)}$

$=\lim_{x\rightarrow 0}\frac{(2x-3)(\sqrt{x}-1)}{(2x+3)((\sqrt{x})^2-1^2)}$

$=\lim_{x\rightarrow 0}\frac{(2x-3)(\sqrt{x}-1)}{(2x+3)(\sqrt{x}^2-1)(\sqrt{x}^2+1)}$

$=\frac{2(1)-3}{(2\times 1+3)(\sqrt{1}+1)}$

$=\frac{2-3}{5\times 2}=\frac{-1}{10}$

Question 30

निकाले (Find)

$\lim_{x\rightarrow 2}\frac{x^2-4}{\sqrt{x+2}-\sqrt{3x-2}}$

Sol :

हर का परिमेय करण करने पर,

$=\lim_{x\rightarrow 2}\frac{x^2-4}{\sqrt{x+2}-\sqrt{3x-2}}\times \frac{\sqrt{x+2}+\sqrt{3x-2}}{\sqrt{x+2}+\sqrt{3x-2}}$

$=\lim_{x\rightarrow 2}\frac{(x^2-4)(\sqrt{x+2}+\sqrt{3x-2})}{(\sqrt{x+2})^2-(\sqrt{3x-2})^2}$

$=\lim_{x\rightarrow 2}\frac{(x^2-2^2)(\sqrt{x+2}+\sqrt{3x-2})}{x+2-3x+2}$

$=\lim_{x\rightarrow 2}\frac{(x-2)(x+2)(\sqrt{x+2}+\sqrt{3x-2})}{-2x+4}$

$=\lim_{x\rightarrow 2}\frac{(x-2)(x+2)(\sqrt{x+2}+\sqrt{3x-2})}{-2(x-2)}$

$=\frac{(2+2)(\sqrt{2+2}+\sqrt{3(2)-2})}{-2}$

$=\frac{4(2+2)}{-2}=\frac{4\times 4}{-2}$

=-8

Question 32

निकाले (Find)

$\lim_{x\rightarrow 0}\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{x^2+1}+\sqrt{1+x}}$

Sol :

$=\frac{\sqrt{1+0^2}-\sqrt{1+0}}{\sqrt{x^2+1}+\sqrt{1+0}}$

$=\frac{1-1}{1+1}=\frac{0}{2}=0$

Question 33

निकाले (Find)

$\frac{\sqrt{1+2x^2}-\sqrt{1+x}}{\sqrt{4+4x^2}-\sqrt{4-3x}}$

Sol :

अंश का परिमेय करण करने पर,

$=\frac{\sqrt{1+2x^2}-\sqrt{1+x}}{\sqrt{4+4x^2}-\sqrt{4-3x}}\times \frac{\sqrt{1+2x^2}+\sqrt{1+x}}{\sqrt{1+2x^2}+\sqrt{1+x}}\times \frac{\sqrt{4+4x^2}+\sqrt{4-3x}}{\sqrt{4+4x^2}+\sqrt{4-3x}}$

$=\lim_{x\rightarrow 0}\frac{\left[(\sqrt{1+2x^2}-(\sqrt{1+x})^2)\right](\sqrt{4+4x^2}+\sqrt{4-3x})}{\left[(\sqrt{4+4x^2})^2-(\sqrt{4-3x})^2\right](\sqrt{1+2x^2}+\sqrt{1+x})}$

$=\lim_{x\rightarrow 0}\frac{(1+2x^2-1-x)(\sqrt{4+4x^2}+\sqrt{4-3x})}{(4+4x^2-4+3x)(\sqrt{1+2x^2}+\sqrt{1+x})}$

$=\lim_{x\rightarrow 0}\frac{x(2x-1)(\sqrt{4+4x^2}+\sqrt{4-3x})}{x(4x+3)(\sqrt{1+2x^2}+\sqrt{1+x})}$

$=\frac{[2(0)-1](\sqrt{4+4(0)^2}+\sqrt{4-3(0)})}{[4(0)+3](\sqrt{1+2(0)^2}+\sqrt{1+0})}$

$=\frac{-1(2+2)}{3(1+1)}$

$=\frac{-1\times 4}{3\times 2}=\frac{-2}{3}$

Question 35

निकाले (Find)

$\lim_{x\rightarrow 1}\frac{x^2-3x+2}{x^2-1}$

Sol :

=\lim_{x\rightarrow 1}\frac{x^2-2x-x+2}{x^2-1^2}$

$=\lim_{x\rightarrow 1}\frac{x(x-2)-1(x-2)}{(x-1)(x+1)}$

$=\lim_{x\rightarrow 1}\frac{(x-2)(x-1)}{(x-1)(x+1)}$

$=\frac{1-2}{1+1}=\frac{-1}{2}$

Question 35

निकाले (Find)

$\lim_{x\rightarrow 1}\frac{x^2-3x+2}{x^2-5x+4}$

Sol :

$=\lim_{x\rightarrow 0}\frac{x^2-2x-x+2}{x^2-4x-x+4}$

$=\lim_{x\rightarrow 1}\frac{x(x-2)-1(x-2)}{x(x-4)-1(x-4)}$

$=\lim_{x\rightarrow 1}\frac{(x-2)(x-1)}{(x-4)(x-1)}$

$=\frac{1-2}{1-4}=\frac{-1}{-3}=\frac{1}{3}$

Question 38

निकाले (Find)

$\lim_{x\rightarrow 2}\left(\dfrac{\frac{1}{x}+\frac{1}{2}}{x+2}\right)$

Sol :

$=\lim_{x\rightarrow -2}\dfrac{\frac{2+x}{2x}}{\frac{x+2}{1}}$

$=\lim_{x\rightarrow -2}\frac{1}{2x}$

$=\frac{1}{2\times -2}=\frac{-1}{4}$

Question 40

निकाले (Find)

$\lim_{x\rightarrow 2}\frac{x^3-2x^2}{x^2-5x+6}$

Sol :

$=\lim_{x\rightarrow 2}\frac{x^2(x-2)}{x^2-3x-2x+6}$

$=\lim_{x\rightarrow 2}\frac{x^2(x-2)}{x(x-3)-2(x-3)}$

$=\lim_{x\rightarrow 2}\frac{x^2(x-2)}{(x-3)(x-2)}$

$=\frac{(2)^2}{2-3}=\frac{4}{-1}$

=-4

Question 41

निकाले (Find)

$\lim_{x\rightarrow 2}\frac{x^3-4x^2+4x}{x^2-4}$

Sol :

$=\lim_{x\rightarrow 2}\frac{x^2-4x+4}{x^2-4}$

$=\lim_{x\rightarrow 2}\frac{x(x^2-2.x.2+2^2)}{(x-2)(x+2)}$

$=\lim_{x\rightarrow 2}\frac{x(x-2)^2}{(x-2)(x+2)}$

$=\frac{2(2-2)}{2+2}$

=0

Question 42

निकाले (Find)

$\lim_{x\rightarrow 3}\frac{x^4-81}{2x^2-5x-3}$

Sol :

$=\lim_{x\rightarrow 3}\frac{(x^2)^2-(3^2)^2}{2x^2-6x+x-3}$

$=\lim_{x\rightarrow 3}\frac{(x^2-3^2)(x^2+3^2)}{2x(x-3)+1(x-3)}$

$=\lim_{x\rightarrow 3}\frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}$

$=\frac{(3+3)(3^2+9)}{2(3)+1}$

$=\frac{6\times 18}{7}=\frac{108}{7}$

Question 43

निकाले (Find)

$\lim_{x\rightarrow 2}\frac{x^2-4}{x^3-4x^2+4x}$

Sol :

$=\lim_{x\rightarrow 2}\frac{x^2-2^2}{x(x^2-4x+4)}$

$=\lim_{x\rightarrow 2}\frac{(x-2)(x+2)}{x(x-2)^2}$

$=\frac{2+2}{2(2-2)}=\frac{4}{2\times 0}=\frac{2}{0}$

Limit does not exist

Question 44

निकाले (Find)

$\lim_{x\rightarrow 2}\frac{2x^3+3x^2-8x-12}{x^3+5x^2-4x-20}$

Sol :

$=\lim_{x\rightarrow 2}\frac{2x^3-4x^2+7x^2-14x+6x-12}{x^3-2x^2+7x^2-14x+10x-20}$

$=\lim_{x\rightarrow 2}\frac{2x^2(x-2)+7x(x-2)+6(x-2)}{x^2(x-2)+7x(x-2)+10(x-2)}$

$=\lim_{x\rightarrow 2}\frac{(x-2)(2x^2+7x+6)}{(x-2)(x^2+7x+10)}$

$=\frac{2(2)^2+7(2)+6}{2^2+7\times 2+10}$

$=\frac{8+14+6}{4+14+10}=\frac{28}{28}$

=1

Question 45

निकाले (Find)

यदि (If) $f(x)=\frac{1}{x}$, तो साबित करे (prove that) $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=-\frac{1}{x^2}$

Sol :

L.H.S

$=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\rightarrow 0}\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}$

$\left[\therefore \lim_{x\rightarrow a}=\frac{x^4-a^4}{x-a}=na^{n-1}\right]$

$=\lim_{h\rightarrow 0}\frac{(x+h)^{-1}-x^{-1}}{(x+h)-x}$

=-1.x^-1-1

=-1.x^-2

$=\frac{-1}{x^2}$

$\lim_{x\rightarrow 2}\left(\frac{1}{x-2}-\frac{4}{x^2-4}\right)$

Sol :

$=\lim_{x\rightarrow 2}\left(\frac{1}{x-2}-\frac{4}{(x-2)(x+2)}\right)$

$=\lim_{x\rightarrow 2}\frac{1}{(x-2)}\left[1-\frac{4}{x+2}\right]$

$=\lim_{x\rightarrow 2}\frac{1}{(x-2)}\left[\frac{x+2-4}{x+2}\right]$

$=\lim_{x\rightarrow 2}\frac{1}{(x-2)}\left[\frac{x-2}{x+2}\right]$

$=\frac{1}{2+2}=\frac{1}{4}$

$\lim_{x\rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2x-3)}{x^3-3x^2+2x}\right]$

Sol :

$=\lim_{x\rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2x-3)}{x(x^2-3x+2)}\right]$

$=\lim_{x\rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2x-3)}{x(x^2-2x-x+2)}\right]$

$=\lim_{x\rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2x-3)}{x[x(x-2)-1(x-2)]}\right]$

$=\lim_{x\rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2x-3)}{x(x-2)(x-1)}\right]$

$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[1-\frac{2(2x-3)}{x(x-1)}\right]$

$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{x(x-1)-2(2x-3)}{x(x-1)}\right]$

$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{x^2-x-4x+6}{x(x-1)}\right]$

$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{x^2-5x+6}{x(x-1)}\right]$

$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{x^2-3x-2x+6}{x(x-1)}\right]$

$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{x(x-3)-2(x-3)}{x(x-1)}\right]$

$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{(x-3)(x-2)}{x(x-1)}\right]$

$=\frac{2-3}{2(2-1)}$

$=\frac{-1}{2(1)}=-\frac{1}{2}$

$\lim_{x\rightarrow 1}\left[\frac{x-2}{x^2-x}-\frac{1}{x^3-3x^2+2x}\right]$

Sol :

$=\lim_{x\rightarrow 1}\left[\frac{x-2}{x(x-1)}-\frac{1}{x(x^2-3x+2)}\right]$

$=\lim_{x\rightarrow 1}\left[\frac{x-2}{x(x-1)}-\frac{1}{x(x^2-2x-x+2)}\right]$

$=\lim_{x\rightarrow 1}\left[\frac{x-2}{x(x-1)}-\frac{1}{x\{x[(x-2)-1(x-2)\}}\right]$

$=\lim_{x\rightarrow 1}\left[\frac{x-2}{x(x-1)}-\frac{1}{x(x-2)(x-1)}\right]$

$=\lim_{x\rightarrow 1} \frac{1}{x(x-1)}\left[x-2-\frac{1}{x-2}\right]$

$=\lim_{x\rightarrow 1} \frac{1}{x(x-1)}\left[\frac{x^2-4x+4-1}{x-2}\right]$

$=\lim_{x\rightarrow 1} \frac{1}{x(x-1)}\left[\frac{x^2-4x+3}{x-2}\right]$

$=\lim_{x\rightarrow 1} \frac{1}{x(x-1)}\left[\frac{x^2-3x-x+3}{x-2}\right]$

$=\lim_{x\rightarrow 1} \frac{1}{x(x-1)}\left[\frac{x(x-3)-1(x-3)}{x-2}\right]$

$=\lim_{x\rightarrow 1} \frac{1}{x(x-1)}\left[\frac{(x-3)(x-1)}{x-2}\right]$

$=\frac{1-3}{1(1-2)}=\frac{-2}{1(-1)}$

$=\frac{-2}{-1}$

=2