Exercise 24.1
Question 1
निम्नलिखित सीमाओ को निकाले(Evaluate the following limits)
(i) $\lim_{x\rightarrow 1}(x-2)$
Sol :
=1-2
=-1
(ii) $\lim_{x\rightarrow 2}(x+2)$
Sol :
=2+2
=4
(iii) $\lim_{x\rightarrow 0}(9)$
Sol :
=9
(iv) $\lim_{x\rightarrow 3} \frac{x^2-9}{x+2}$
Sol :
$=\frac{3^2-9}{3+2}$
$=\frac{0}{5}=0$
(v) $\lim_{x\rightarrow 2}(3-x)$
Sol :
(vi) $\lim_{x\rightarrow 2}x(x-1)$
(vii) $\lim_{x\rightarrow -1} \frac{x^3-3x+1}{x-1}$
(viii) $\lim_{x\rightarrow 2} \frac{x^2+4}{x+2}$
(ix) $\lim_{x\rightarrow 1}\frac{x^2+5}{x-2}$
Sol
$=\frac{1^2+5}{1-2}$ :
$=\frac{a(0)+b}{c(0)+d}$
$=\frac{b}{a}$
(x) $\lim_{x\rightarrow 0}\frac{ax+b}{cx+d}$,d≠0
Sol :
$=\frac{a(0)+b}{c(0)+d}=\frac{b}{a}$
(xi)
$\lim_{x\rightarrow 0} \frac{3x+1}{x+3}$
Sol :
$=\frac{6}{-1}=-6$
(xii) $\lim_{x\rightarrow 0} \frac{x^2-4x}{x-2}$
(xiii) $\lim_{r\rightarrow 1} \pi r^2$
(xiv) $\lim_{x\rightarrow 3} x(x+3)$
(xv) $\lim_{x\rightarrow 3} x+3$
(xvi) $\lim_{x\rightarrow 1} (x^3-x^2+1)$
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.a4-1
=4a3
Sol :
Sol :
$\lim_{x\rightarrow a}\frac{x^n-a^n}{x-a}=n.a^{n-1}$
=4.a4-1
=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
अंश तथा हर मे x-(3) से भाग देने पर
$\lim_{x\rightarrow -3}\dfrac{\frac{x^3-(-3)^3}{x-(-3)}}
{\frac{x^5-(-3)^5}{x-(-3)}}$
$=\dfrac{\lim_{x\rightarrow -3}\frac{x^3-(-3)^3}{\lim_{x\rightarrow
-3}}}{\frac{x^5-(-3)^5}{x-(-3)}}$
$=\frac{3(-3)^{3-1}}{5(-3)^{5-1}}$
$=\frac{3(-3)^2}{5(-3)^4}$
$=\frac{3\times 9}{5\times 81}=\frac{1}{15}$
Sol :
$=\frac{3(-3)^2}{5(-3)^4}$
$=\frac{3\times 9}{5\times 81}=\frac{1}{15}$
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 :
Question 7
निकाले
(Find)
$\lim_{x\rightarrow a}\frac{x-a}{x^{3/2}-a^{3/2}}$
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
निकाले
(Find)
$\lim_{x\rightarrow 2}\frac{\sqrt[3]{x-1}-1}{x-2}$
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
निकाले
(Find)
$\lim_{x\rightarrow 0}\frac{(1+x)^n-1}{x}$
Sol :
$=\lim_{x\rightarrow 0}\frac{(1+x)^n-1^n}{(1+x)-1}$
=n(1)n-1
=n
Question 10
निकाले (Find)
$\lim_{x\rightarrow 0}\frac{(1+x)^{1/3}-(1-x)^{1/3}}{x}$
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
निकाले (Find)
$\lim_{x\rightarrow 1}\left(\frac{1-x^{-1/3}}{1-x^{-2/3}}\right)$
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.a3-1=4.14-1
3a2=4
a2=$\frac{4}{3}$
$a=\pm \sqrt{\frac{4}{3}}$
$a=\pm \frac{2}{\sqrt{3}}$
Question 20
निकाले (Find)
$\lim_{x\rightarrow 0}\frac{\sqrt{1+4x}-\sqrt{1-3x}}{x}$
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
$=\frac{-1}{x^2}$
Question 46
निकाले (Find)
$\lim_{x\rightarrow 2}\left(\frac{1}{x-2}-\frac{4}{x^2-4}\right)$
Sol :
$=\lim_{x\rightarrow 2}\left(\frac{1}{x-2}-\frac{4}{(x-2)(x+2)}\right)$
$=\lim_{x\rightarrow 2}\frac{1}{(x-2)}\left[1-\frac{4}{x+2}\right]$
$=\lim_{x\rightarrow 2}\frac{1}{(x-2)}\left[\frac{x+2-4}{x+2}\right]$
$=\lim_{x\rightarrow 2}\frac{1}{(x-2)}\left[\frac{x-2}{x+2}\right]$
$=\frac{1}{2+2}=\frac{1}{4}$
Question 47
निकाले (Find)
$\lim_{x\rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2x-3)}{x^3-3x^2+2x}\right]$
Sol :
$=\lim_{x\rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2x-3)}{x(x^2-3x+2)}\right]$
$=\lim_{x\rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2x-3)}{x(x^2-2x-x+2)}\right]$
$=\lim_{x\rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2x-3)}{x[x(x-2)-1(x-2)]}\right]$
$=\lim_{x\rightarrow 2}\left[\frac{1}{x-2}-\frac{2(2x-3)}{x(x-2)(x-1)}\right]$
$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[1-\frac{2(2x-3)}{x(x-1)}\right]$
$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{x(x-1)-2(2x-3)}{x(x-1)}\right]$
$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{x^2-x-4x+6}{x(x-1)}\right]$
$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{x^2-5x+6}{x(x-1)}\right]$
$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{x^2-3x-2x+6}{x(x-1)}\right]$
$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{x(x-3)-2(x-3)}{x(x-1)}\right]$
$=\lim_{x\rightarrow 2} \frac{1}{(x-2)}\left[\frac{(x-3)(x-2)}{x(x-1)}\right]$
$=\frac{2-3}{2(2-1)}$
$=\frac{-1}{2(1)}=-\frac{1}{2}$
Question 48
निकाले (Find)
$\lim_{x\rightarrow 1}\left[\frac{x-2}{x^2-x}-\frac{1}{x^3-3x^2+2x}\right]$
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
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