KC Sinha Solution Class 11 Chapter 25 सीमा का अस्तित्व (Existence of Limit) Exercise 25.1

 Exercise 25.1

Question 1

यदि (If) f(x)=x²+x+2, जब(when) x<1
=x⁴+3,जब(when)x>1

तो क्या $\lim _{x \rightarrow 1} f(x)$ का अस्तित्व है।(then does $\lim _{x \rightarrow 1} f(x)$ exist?)

Sol :

L.H.L

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}\left(x^{2}+x+2\right)$

=1²+1+2

=4

R.H.L

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(x^{4}+3\right)$

=1⁴+3

=4

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=4$

$\lim _{x \rightarrow 1} f(x)$ exist

$\lim _{x \rightarrow 1} f(x)=4$

Question 2

यदि (If) f(x)=x³,जब(when)x<-1
=x⁵,जब(when) x>-1
=-1, जब(when) x=-1

तो $\lim _{x \rightarrow-1} f(x)$ ज्ञात करे।(then find $\lim _{x \rightarrow-1} f(x)$)

Sol :

L.H.L

$\lim _{x \rightarrow-1^{-}} f(2)=\lim _{x \rightarrow-1}\left(x^{3}\right)$

=(-1)³

=-1

R.H.L

$\lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1}\left(x^{5}\right)$

=(-1)⁵

=-1

$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{+}} f(x)=-1$

$\lim _{x \rightarrow-1} f(x)$ exist

$\lim _{x \rightarrow-1} f(x)=-1$

Question 3

ज्ञात करे(Find) $\lim _{x \rightarrow 2} \phi(x)$,जहाँ (where)
ф(x)=5; जब(when) x≠2
=3,(when)x=2

Sol :

L.H.L

$\lim _{x \rightarrow 2^{-}} \phi(x)=\lim _{x \rightarrow 2}(5)=5$

R.H.L

$=\lim_{x\rightarrow 2^+}\phi(x)=\lim _{x \rightarrow 2} 5=5$

$\lim _{x \rightarrow 2^-} \phi(x)=\lim _{x \rightarrow 2^{+}} \phi(x)=5$

$\lim _{x \rightarrow 2} \phi(x)$ exist

$\lim _{x \rightarrow 2} \phi(x)=5$

Question 4

(i) $\lim _{x \rightarrow 0} f(x)$, ज्ञात कीजिए

Evaluate $\lim _{x \rightarrow 0} f(x)$, जहाँ(where) $f(x)=\left\{\begin{aligned} \frac{x}{|x|}, & x \neq 0 \\ 0, & x=0 \end{aligned}\right.$

Sol :

L.H.L

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}} \frac{x}{|x|}$

$=\lim _{x \rightarrow 0} \frac{x}{-x}=-1$

R.H.L

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}} \frac{x}{|2|}$

$=\lim _{x \rightarrow 0} \frac{x}{x}=1$

$\lim _{x \rightarrow 0^{-}} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$

$\lim _{x \rightarrow 0} f(x)$ is not exist

Question 5

साबित करे कि (prove that) $\lim _{x \rightarrow 1} \frac{x^{2}-1}{|x-1|}$ का अस्तित्व नही है।(does not exist)

Sol :

L.H.L

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}} \frac{x^{2}-1}{|x-1|}$

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

$=\lim _{x \rightarrow 1} \frac{(x-1)(x+1)}{-(x-1)}$

=-(1+1)=-2

R.H.L

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1^{+}} \frac{x^{2}-1}{|x-1|}$

$=\lim _{x \rightarrow 1} \frac{x^{2}-1^{2}}{x-1}$

$=\lim _{x \rightarrow 1} \frac{(x -1)(x+1)}{x-1}$

=1+1

=2

$\lim_{x\rightarrow 1^-}f(x) \neq \lim_{x\rightarrow 1^+}f(x)$

अतः $\lim _{x \rightarrow 1} f(x)$ का अस्तित्व नही है।

Question 6

यदि (If) $f(x)=\left\{\begin{array}{cc}4, & x \geq 3 \\ x+1, & x<3\end{array}\right.$

ज्ञात कीजिए (Find) $\lim _{x \rightarrow 3} f(x)$

Sol :

L.H.L

$\lim _{x \rightarrow 3^{-}} f(x)=\lim _{x \rightarrow 3}(x+1)$

=3+1

=4

R.H.L

$\lim _{x \rightarrow 3^{+}} f(x)=\lim _{x \rightarrow 3}(4)=4$

$\lim_{x \rightarrow 3^-} f(x)=\lim _{x \rightarrow 3^{+}} f(x)=4$

$\lim _{x \rightarrow 3} f(x)$ is exist

$\lim _{x \rightarrow 3} f(x)=4$

Question 8

$\lim _{x \rightarrow 5^{+}} f(x)$ तथा $\lim _{x \rightarrow 5^{-}} f(x)$ निकाले, जहाँ f(x)=|x|-5

(Find $\lim _{x \rightarrow 5^{+}} f(x)$ and $\lim _{x \rightarrow 5^{-}} f(x)$ where f(x)=|x|-5)

Sol :

L.H.L

$\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{-}}|x|-5$

$=\lim _{x \rightarrow 5}(x-5)$

=5-5

=0

R.H.L

$\lim _{x \rightarrow 5^{+}} f(x)=\lim _{x \rightarrow 5^{+}}|x|-5$

$=\lim _{x \rightarrow 5}(x-5)$

=5-5

=0

Question 9

$\lim _{x \rightarrow 0} f(x)$ और $\lim _{x \rightarrow 1} f(x)$ ज्ञात कीजिए, जहाँ $f(x)=\left\{\begin{array}{ll}2 x+3, & x \leq 0 \\ 3(x+1), & x>0\end{array}\right.$

Sol :

At x=0

L.H.L

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}(2 x+3)$

=2(0)+3

=3

R.H.L

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} 3(x+1)$

=3(0+1)

=3

$\lim_{x \rightarrow 0^-} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=3$

$\lim _{x \rightarrow 0} f(x)$ is exist

$\lim_{x\rightarrow 0}=f(x)=3$

At x=1

L.H.L

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} 3(x+1)$

=3(1+1)

=3(2)

=6

R.H.L

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 3(x+1)$

=3(1+1)

=6

$\lim_{x\rightarrow1^-}f(x)=\lim _{x \rightarrow 1^+}f(x)=6$

$\lim _{x \rightarrow 1} f(x)$ is exist

$\lim _{x \rightarrow 1} f(x)=6$

Question 10

निम्नलिखित सीमाओ को निकाले यदि उनका अस्तित्व है।

[Find the following limits if they exist]

(i) $\lim _{x \rightarrow 1}[x]$

Sol :

L.H.L

$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{-}}[x]=0$

R.H.L

$\lim_{x\rightarrow 1^+}f(x)=\lim _{x \rightarrow 1^+}[x]=1$

$\lim_{x\rightarrow 1^-}f(x)\neq\lim_{x\rightarrow 1^+}f(x)$

अतः $\lim _{x \rightarrow 1}[x]$ का अस्तित्व नही है।

(ii) $\lim _{x \rightarrow \frac{5}{2}}[x]$

Sol :

L.H.L

$\lim _{x \rightarrow \frac{5}{2}^+}f(x)=\lim _{x \rightarrow \frac{5}{2}^-}[x]=2$

R.H.L

$\lim _{x \rightarrow \frac{5}{2}^+}f(x)=\lim_{x\rightarrow \frac{5}{2}^+}[x]=2$

$\lim _{x \rightarrow \frac{5}{2}^-}[x]=\lim _{x \rightarrow \frac{5}{2}^+}[x]$

अतः $\lim _{x \rightarrow \frac{5}{2}}[x]$ is exist

$\lim _{x \rightarrow \frac{5}{2}}[x]=2$

(iii) $\lim _{x \rightarrow \frac{7}{3}}[-x]$

Sol :

L.H.L

$\lim _{x \rightarrow \frac{7}{3}^-}[-x]=\lim _{x \rightarrow \frac{7}{3}^-}[-x]=-3$

R.H.L

$\lim _{x \rightarrow \frac{7}{3}^+}[-x]=-3$

$\lim _{x \rightarrow \frac{7}{3}^{-}}[-x]=\lim _{x \rightarrow \frac{7}{3}^+}[-x]=-3$

अतः $\lim _{x \rightarrow \frac{2}{3}}[-x]$ का अस्तित्व है।

$\lim _{x \rightarrow \frac{7}{3}}[-x]=-3$

Question 11

(i) ज्ञात करे(Find) $\lim _{x \rightarrow-1-0} \frac{\sqrt{x^{2}-5 x-6}}{x+3}$

(ii) ज्ञात करे(Find) $\lim _{x \rightarrow 2^{-}} \frac{x^{2}-4 x+4}{x-2}$

Sol :

(i) $\lim _{x \rightarrow-1^{-}} \frac{\sqrt{x^{2}-5 x-6}}{x+3}$

$=\frac{\sqrt{(-1)^{2}-5(-1)-6}}{-1+3}$

$=\frac{\sqrt{1+5-6}}{2}=\frac{0}{2}$

=0

(ii) $\lim _{x \rightarrow 2^{-}} \frac{x^{2}-4 x+4}{x-2}$

$=\lim _{x \rightarrow 2^{-}} \frac{x^{2}-2 \cdot x \cdot 2+2^{2}}{x-2}$

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

=2-2

=0

Question 12

यदि (If) $f(x)=\frac{\sin 3 x}{x}$, जहाँ (where)x<0

$=\frac{\tan b x}{x}$, जब (when)x>0

तथा $\lim _{x \rightarrow 0} f(x)$ का अस्तित्व है तो b का मान निकाले।

(and $\lim _{x \rightarrow 0} f(x)$ exists, find the value of b)

Sol :

L.H.L

$\lim_{x\rightarrow 0^-}f(x)=\lim _{x \rightarrow 0} \frac{\sin 3 x}{3 x} \times 3$

=1×3

=3

R.H.L

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} \frac{\tan b x}{b x} \times b$

=1×6

=6

$\because \lim _{x \rightarrow 0} f(x)$ का अस्तित्व है।

$\lim_{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)$

3=b

Question 13

माना कि(Let) $f(x)=\left\{\begin{array}{ll}x+2, & x \leq-1 \\ c x^{2}, & x>-1\end{array}\right.$

c का मान निकाले यदि $\lim _{x \rightarrow-1} f(x)$ का अस्तित्व है।

(Find c if $\lim _{x \rightarrow-1} f(x)$ exists)

Sol :

L.H.L

$\lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1}(x+2)$

=-1+2

=1

R.H.L

$\lim_{x\rightarrow {-1}^+}f(x)=\lim _{x \rightarrow-1}cx^{2}$

=c(-1)²

=c

∵$\lim_{x\rightarrow -1} f(x)$ का अस्तित्व है।

$\lim _{x \rightarrow-1^-} f(x)=\lim _{x \rightarrow-1^{+}} f(x)$

1=c