MCQ 1512 Marks
$\lim _{x \rightarrow 2} \frac{x^2-4}{x \sqrt{x}-2 \sqrt{2}}=$
- A
$\frac{4}{3} \sqrt{3}$
- B
$\frac{4}{2} \sqrt{3}$
- ✓
$\frac{4}{3} \sqrt{2}$
- D
$\frac{4}{2} \sqrt{2}$
AnswerCorrect option: C. $\frac{4}{3} \sqrt{2}$
(C)
$\lim _{x \rightarrow 2} \frac{x^2-4}{x \sqrt{x}-2 \sqrt{2}}=\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(x \sqrt{x}+2 \sqrt{2})}{(x \sqrt{x}-2 \sqrt{2})(x \sqrt{x}+2 \sqrt{2})}$
$=\lim _{x \rightarrow 2} \frac{\left(x^2-4\right)(x \sqrt{x}+2 \sqrt{2})}{x^3-2^3}$
$=\lim _{x \rightarrow 2} \frac{(x+2)(x \sqrt{x}+2 \sqrt{2})}{x^2+2 x+4}$
$=\frac{(2+2)(2 \sqrt{2}+2 \sqrt{2})}{(2)^2+2(2)+4}=\frac{4}{3} \sqrt{2}$
Alternate method:
Apply L-Hospital's Rule.
View full question & answer→MCQ 1522 Marks
$\lim _{x \rightarrow 0} \frac{\sqrt{a+x}-\sqrt{a-x}}{x \sqrt{a(a+x)}}=$
- A
$\frac{1}{\sqrt{a}}$
- B
$\frac{1}{a}$
- C
$\sqrt{ a }$
- ✓
$\frac{1}{a \sqrt{a}}$
AnswerCorrect option: D. $\frac{1}{a \sqrt{a}}$
(D)
$\lim _{x \rightarrow 0} \frac{\sqrt{ a +x}-\sqrt{ a -x}}{x \sqrt{ a ( a +x)}}$
$=\lim _{x \rightarrow 0} \frac{(\sqrt{ a +x}-\sqrt{ a -x})(\sqrt{ a +x}+\sqrt{ a -x})}{x \sqrt{ a ( a +x)}(\sqrt{ a +x}+\sqrt{ a -x})}$
$=\lim _{x \rightarrow 0} \frac{2}{\sqrt{a(a+x)}(\sqrt{a+x}+\sqrt{a-x})}$
$=\frac{2}{\sqrt{a(a+0)}(\sqrt{a+0}+\sqrt{a-0})}=\frac{1}{a \sqrt{a}}$
View full question & answer→MCQ 1532 Marks
The value of $\lim _{x \rightarrow 7} \frac{2-\sqrt{x-3}}{x^2-49}$ is
- A
$\frac{2}{9}$
- B
$-\frac{2}{49}$
- C
$\frac{1}{56}$
- ✓
$-\frac{1}{56}$
AnswerCorrect option: D. $-\frac{1}{56}$
(D)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 7} \frac{2-\sqrt{x-3}}{x^2-49}=\lim _{x \rightarrow 7} \frac{0-\frac{1}{2 \sqrt{x-3}}}{2 x}$
$=\lim _{x \rightarrow 7} \frac{-1}{4 x \sqrt{x-3}}=\frac{-1}{4(7)(2)}$
$=\frac{-1}{56}$
View full question & answer→MCQ 1542 Marks
$\lim _{x \rightarrow 3} \frac{\sqrt{x^2+10}-\sqrt{19}}{x-3}$ is equal to
- A
$0$
- ✓
$\frac{3}{\sqrt{19}}$
- C
$\frac{6}{\sqrt{19}}$
- D
$\sqrt{19}$
AnswerCorrect option: B. $\frac{3}{\sqrt{19}}$
(B)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 3} \frac{\sqrt{x^2+10} \sqrt{19}}{x-3}=\lim _{x \rightarrow 3} \frac{1}{2 \sqrt{x^2+10}}(2 x)=\frac{3}{\sqrt{19}}$
View full question & answer→MCQ 1552 Marks
$\lim _{x \rightarrow-3} \frac{\sqrt{x^2+7}-4}{x+3}$ is equal to
- ✓
$-\frac{3}{4}$
- B
$\frac{3}{4}$
- C
$\frac{3}{9}$
- D
$\frac{-3}{9}$
AnswerCorrect option: A. $-\frac{3}{4}$
(A)
Applying L-Hospital's Rule, we get
$\lim _{x \rightarrow-3} \frac{\sqrt{x^2+7}-4}{x+3}=\lim _{x \rightarrow-3} \frac{1}{2 \sqrt{x^2+7}}(2 x)=\frac{-3}{4}$
View full question & answer→MCQ 1562 Marks
$\lim _{x \rightarrow-2} \frac{\sqrt{x^2+5}-3}{x+2}$ is equal to
- ✓
$-\frac{2}{3}$
- B
$0$
- C
$\frac{2}{3}$
- D
$\frac{2}{9}$
AnswerCorrect option: A. $-\frac{2}{3}$
(A)
Applying L-Hospital's Rule, we get
$\lim _{x \rightarrow-2} \frac{\sqrt{x^2+5}-3}{x+2}=\lim _{x \rightarrow-2} \frac{\frac{1}{2 \sqrt{x^2+5}}(2 x)}{1}=\frac{-2}{3}$
View full question & answer→MCQ 1572 Marks
$\lim _{x \rightarrow 1} \frac{x^8-2 x+1}{x^4-2 x+1}$ equals
Answer(A)
Applying L-Hospital's Rule, we get
$\lim _{x \rightarrow 1} \frac{x^8-2 x+1}{x^4-2 x+1}=\lim _{x \rightarrow 1} \frac{8 x^7-2}{4 x^3-2}=3$
View full question & answer→MCQ 1582 Marks
$\lim _{x \rightarrow 2} \frac{x^6-24 x-16}{x^3+2 x-12}=$
Answer(C)
By synthetic division,
$x^6-24 x-16$
$=(x-2)\left(x^5+2 x^4+4 x^3+8 x^2+16 x+8\right)$
and $x^3+2 x-12=(x-2)\left(x^2+2 x+6\right)$
$\therefore \quad \lim _{x \rightarrow 2} \frac{x^6-24 x-16}{x^3+2 x-12}$
$=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^5+2 x^4+4 x^3+8 x^2+16 x+8\right)}{(x-2)\left(x^2+2 x+6\right)}$
$=\lim _{x \rightarrow 2} \frac{x^5+2 x^4+4 x^3+8 x^2+16 x+8}{x^2+2 x+6}$
$=\frac{168}{14}=12$
Alternate method:
Apply L-Hospital's Rule.
View full question & answer→MCQ 1592 Marks
$\lim _{x \rightarrow 2} \frac{x^4-4 x^3+8 x^2-16 x+16}{x^3-3 x^2+4}=$
- ✓
$\frac{8}{3}$
- B
$\frac{4}{3}$
- C
$\frac{7}{2}$
- D
$\frac{5}{2}$
AnswerCorrect option: A. $\frac{8}{3}$
(A)
$\lim _{x \rightarrow 2} \frac{x^4-4 x^3+8 x^2-16 x+16}{x^3-3 x^2+4}$
$=\lim _{x \rightarrow 2} \frac{(x-2)^2\left(x^2+4\right)}{(x-2)^2(x+1)}=\lim _{x \rightarrow 2} \frac{x^2+4}{x+1}=\frac{8}{3}$
View full question & answer→MCQ 1602 Marks
$\lim _{x \rightarrow 2} \frac{x+x^2+x^3-14}{x-2}=$
Answer(A)
Applying L-Hospital's Rule, we get
$\lim _{x \rightarrow 2} \frac{x+x^2+x^3-14}{x-2}=\lim _{x \rightarrow 2}\left(1+2 x+3 x^2\right)=17$
View full question & answer→MCQ 1612 Marks
The value of the limit of $\frac{x^3-x^2-18}{x-3}$ as $x$ tends to 3 is
Answer(D)
Let $y=\lim _{x \rightarrow 3} \frac{x^3-x^2-18}{x-3}$
Applying L-Hospital's rule, we get
$y=\lim _{x \rightarrow 3}\left(3 x^2-2 x\right)=(27-6)=21$
View full question & answer→MCQ 1622 Marks
$\lim _{x \rightarrow 9}\left(\frac{x^{3 / 2}-27}{x-9}\right)=$
- A
$\frac{3}{2}$
- ✓
$\frac{9}{2}$
- C
$\frac{2}{3}$
- D
$\frac{1}{3}$
AnswerCorrect option: B. $\frac{9}{2}$
(B)
$\lim _{x \rightarrow 9}\left(\frac{x^{3 / 2}-27}{x-9}\right)$
$=\lim _{x \rightarrow 9}\left(\frac{(\sqrt{x})^3-3^3}{(\sqrt{x}+3)(\sqrt{x}-3)}\right)$
$=\lim _{x \rightarrow 9}\left(\frac{(\sqrt{x}-3)(x+3 \sqrt{x}+9)}{(\sqrt{x}+3)(\sqrt{x}-3)}\right)$
$=\frac{9+3 \sqrt{9}+9}{\sqrt{9}+3}$
$=\frac{27}{6}$
$=\frac{9}{2}$
View full question & answer→MCQ 1632 Marks
$\lim _{x \rightarrow 1} \frac{1-x^{\frac{-2}{3}}}{1-x^{\frac{1}{3}}}$ is equal to
- ✓
- B
- C
$\frac{2}{3}$
- D
$\frac{1}{3}$
AnswerA)
$\lim _{x \rightarrow 1} \frac{1-x^{-\frac{2}{3}}}{1-x^{-\frac{1}{3}}}=\lim _{x \rightarrow 1} \frac{1^2-\left(x^{-\frac{1}{3}}\right)^2}{1-x^{\frac{1}{3}}}$
$=\lim _{x \rightarrow 1} \frac{\left(1+x^{-\frac{1}{3}}\right)\left(1-x^{-\frac{1}{3}}\right)}{1-x^{-\frac{1}{3}}}$
$=\lim _{x \rightarrow 1}\left(1+x^{-1 / 3}\right)=2$
Alternate method:
Apply L- Hospital's Rule.
View full question & answer→MCQ 1642 Marks
$\lim _{y \rightarrow 1}\left(\frac{1}{y^2-1}-\frac{2}{y^4-1}\right)=$
- ✓
$\frac{1}{2}$
- B
$\frac{1}{3}$
- C
$\frac{1}{4}$
- D
$0$
AnswerCorrect option: A. $\frac{1}{2}$
(A)
$\lim _{y \rightarrow 1}\left(\frac{1}{y^2-1}-\frac{2}{y^4-1}\right)$
$=\lim _{y \rightarrow 1}\left[\frac{1}{y^2-1}-\frac{2}{\left(y^2+1\right)\left(y^2-1\right)}\right]$
$=\lim _{y \rightarrow 1} \frac{y^2+1-2}{\left(y^2+1\right)\left(y^2-1\right)}$
$=\lim _{y \rightarrow 1} \frac{1}{y^2+1}$
$=\frac{1}{2}$
View full question & answer→MCQ 1652 Marks
$\lim _{x \rightarrow 1}\left(\frac{2}{1-x^2}+\frac{1}{x-1}\right)=$
- ✓
$\frac{1}{2}$
- B
$\frac{1}{5}$
- C
$\frac{1}{7}$
- D
$\frac{1}{11}$
AnswerCorrect option: A. $\frac{1}{2}$
(A)
$\lim _{x \rightarrow 1}\left(\frac{2}{1-x^2}+\frac{1}{x-1}\right)=\lim _{x \rightarrow 1}\left[\frac{2}{(1+x)(1-x)}+\frac{1}{x-1}\right]$
$=\lim _{x \rightarrow 1}\left[\frac{1-x}{(1+x)(1-x)}\right]=\frac{1}{2}$
View full question & answer→MCQ 1662 Marks
If $f: R \rightarrow R$ is defined by $f(x)=[x-3]+|x-4|$ for $x \in R$, then $\lim _{x \rightarrow 3^{-}} f(x)$ is equal to
Answer(C)
$\lim _{x \rightarrow 3^{-}} f (x)=\lim _{x \rightarrow 3^{-}}[x-3]+\lim _{x \rightarrow 3^{-}}|x-4|$
$=\lim _{h \rightarrow 0}[3-h-3]+\lim _{h \rightarrow 0}|3-h-4|$
$=\lim _{ h \rightarrow 0}[- h ]+\lim _{ h \rightarrow 0}|-1- h |$
$=\lim _{ h \rightarrow 0}(-1)+\lim _{ h \rightarrow 0}(1+ h )$
$=-1+1=0$
View full question & answer→MCQ 1672 Marks
$\lim _{x \rightarrow 1}[x]=$
Answer(C)
$\lim _{x \rightarrow 1^{-}}[x]=\lim _{h \rightarrow 0}[1-h]=\lim _{h \rightarrow 0} 0=0$
and $\lim _{x \rightarrow 1^{+}}[x]=\lim _{h \rightarrow 0}[1+h]=\lim _{h \rightarrow 0} 1=1$
Hence, limit does not exist.
View full question & answer→MCQ 1682 Marks
$\lim _{x \rightarrow 2} \frac{|x-2|}{x-2}=$
Answer(C)
$\lim _{x \rightarrow 2^{-}} \frac{|x-2|}{x-2}=\lim _{ h \rightarrow 0} \frac{|2- h -2|}{2- h -2}=-1$
and $\lim _{x \rightarrow 2^{+}} \frac{|x-2|}{x-2}=\lim _{h \rightarrow 0} \frac{|2+h-2|}{2+h-2}=1$
Hence, limit does not exist.
View full question & answer→MCQ 1692 Marks
If the function $f (x)$ satisfies $\lim _{x \rightarrow 1} \frac{ f (x)-3}{x^2-1}=\pi$, then $\lim _{x \rightarrow 1} f(x)$ is
Answer(C)
$\lim _{x \rightarrow 1} \frac{ f (x)-3}{x^2-1}=\pi$
$\Rightarrow \lim _{x \rightarrow 1} f (x)-\lim _{x \rightarrow 1} 3=\pi \lim _{x \rightarrow 1}\left(x^2-1\right)$
$\Rightarrow \lim _{x \rightarrow 1} f(x)-3=\pi(0)$
$\Rightarrow \lim _{x \rightarrow 1} f (x)-3=0$
$\Rightarrow \lim _{x \rightarrow 1} f (x)=3$
View full question & answer→MCQ 1702 Marks
If $\lim _{x \rightarrow 1} \frac{x+x^2+x^3+\ldots+x^n- n }{x-1}=5050$, then $n =$
Answer(B)
$\lim _{x \rightarrow 1} \frac{x+x^2+x^3+\ldots+x^{ n }- n }{x-1}=5050$
$\therefore \quad \frac{ n ( n +1)}{2}=5050 \Rightarrow n =100$
View full question & answer→MCQ 1712 Marks
The value of $\lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^2-n}{x-1}$ is
- A
- B
$\frac{n+1}{2}$
- ✓
$\frac{n(n+1)}{2}$
- D
$\frac{ n ( n -1)}{2}$
AnswerCorrect option: C. $\frac{n(n+1)}{2}$
(C)
$\lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^{ n }- n }{x-1}$
$=\lim _{x \rightarrow 1} \frac{(x-1)+\left(x^2-1^2\right)+\left(x^3-1^3\right)+\ldots+\left(x^{ n }-1^{ n }\right)}{x-1}$
$=\lim _{x \rightarrow 1}\left(\frac{x-1}{x-1}+\frac{x^2-1^2}{x-1}+\frac{x^3-1^3}{x-1}+\ldots+\frac{x^{ n }-1^{ n }}{x-1}\right)$
$=1+2(1)^{2-1}+3(1)^{3-1}+\ldots+ n (1)^{ n -1}$
$=1+2+3+\ldots+n$
$=\frac{ n ( n +1)}{2}$
Alternate method:
Applying L-Hospital's Rule, we get
$\lim _{x \rightarrow 1}\left(1+2 x+\ldots \ldots .+n x^{n-1}\right)$
$=1+2+3+\ldots+n=\frac{n(n+1)}{2}$
View full question & answer→MCQ 1722 Marks
If $\lim _{x \rightarrow-a} \frac{x^9+a^9}{x+a}=9$, then $a=$
- ✓
$\pm 1$
- B
$\pm 3$
- C
$\pm 2$
- D
$\pm 4$
AnswerCorrect option: A. $\pm 1$
(A)
$\lim _{x \rightarrow- a } \frac{x^9+ a ^9}{x+ a }=9$
$\Rightarrow \lim _{x \rightarrow- a } \frac{x^9-(- a )^9}{x-(- a )}=9$
$\Rightarrow 9(-a)^{9-1}=9 \Rightarrow a= \pm 1$
View full question & answer→MCQ 1732 Marks
If $\lim _{x \rightarrow 5} \frac{x^4-5^4}{x-5}=500$, then the positive integral value of $k$ is
Answer(B)
We know that, $\lim _{x \rightarrow a } \frac{x^{ n }- a ^{ n }}{x- a }= n a ^{ n -1}$
$\therefore \quad \lim _{x \rightarrow 5} \frac{x^{ k }-5^{ k }}{x-5}= k (5)^{ k -1}$
Given, $\lim _{x \rightarrow 5} \frac{x^{ k }-5^{ k }}{x-5}=500$
$\therefore \quad k (5)^{ k -1}=500$
$\Rightarrow k (5)^{ k -1}=4(5)^{4-1}$
$\Rightarrow k=4$
View full question & answer→MCQ 1742 Marks
If $\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow k} \frac{x^3-k^3}{x^2-k^2}$, then $k=$
- A
$\frac{4}{3}$
- ✓
$\frac{8}{3}$
- C
$\frac{2}{3}$
- D
$\frac{3}{8}$
AnswerCorrect option: B. $\frac{8}{3}$
(B)
$\lim _{x \rightarrow 1} \frac{x^4-1}{x-1}=\lim _{x \rightarrow 1} \frac{x^4-1^4}{x-1}=4$
$\lim _{x \rightarrow k } \frac{x^3- k ^3}{x^2- k ^2}=\frac{3}{2} k ^{3-2}=\frac{3}{2} k$
$\therefore \quad 4=\frac{3 k }{2} \Rightarrow k =\frac{8}{3}$
View full question & answer→MCQ 1752 Marks
$\lim _{x \rightarrow 2} \frac{x^{\frac{7}{2}}-2^{\frac{7}{2}}}{x^{\frac{3}{2}}-2^{\frac{3}{2}}}=$
- ✓
$\frac{28}{3}$
- B
$\frac{34}{3}$
- C
$\frac{29}{2}$
- D
$\frac{35}{3}$
AnswerCorrect option: A. $\frac{28}{3}$
(A)
$\lim _{x \rightarrow a} \frac{x^m-a^m}{x^n-a^n}=\frac{m}{n} a^{m-n}$
$\lim _{x \rightarrow 2} \frac{x^{\frac{7}{2}}-2^{\frac{7}{2}}}{x^{\frac{3}{2}}-2^{\frac{3}{2}}}=\frac{7}{2} \times \frac{2}{3}(2)^{\frac{7}{2}-\frac{3}{2}}=\frac{7}{3} \times 4=\frac{28}{3}$
View full question & answer→MCQ 1762 Marks
$\lim _{x \rightarrow a} \frac{x^{-5}-a^{-5}}{x^{-3}-a^{-3}}=$
- A
$\frac{9}{4 a^2}$
- B
$-\frac{5}{3 a^2}$
- ✓
$\frac{5}{3 a^2}$
- D
$-\frac{9}{4 a^2}$
AnswerCorrect option: C. $\frac{5}{3 a^2}$
(C)
$\lim _{x \rightarrow a} \frac{x^m-a^m}{x^n-a^n}=\frac{m}{n} a^{m-n}$
$\lim _{x \rightarrow a} \frac{x^{-5}-a^{-5}}{x^{-3}-a^{-3}}=\frac{-5}{-3}$(a) ${ }^{-5-(-3)}=\frac{5}{3}$$(a)^{-2}=\frac{5}{3 a^2}$
View full question & answer→MCQ 1772 Marks
$\lim _{x \rightarrow 1} \frac{x^{100}-1}{x^{50}-1}=$
Answer(D)
$\lim _{x \rightarrow a} \frac{x^m-a^m}{x^n-a^n}=\frac{m}{n} a^{m-n}$
$\lim _{x \rightarrow 1} \frac{x^{100}-1}{x^{50}-1}=\frac{100}{50}(1)^{100-50}=2$
View full question & answer→MCQ 1782 Marks
The value of $\lim _{x \rightarrow 3} \frac{x^5-243}{x^2-9}$ is
AnswerCorrect option: C. $\frac{135}{2}$
(C)
$\lim _{x \rightarrow a} \frac{x^m-a^m}{x^n-a^n}=\frac{m}{n} a^{m-n}$
$\lim _{x \rightarrow 3} \frac{x^5-243}{x^2-9}=\lim _{x \rightarrow 3} \frac{x^5-3^5}{x^2-3^2}=\frac{5}{2}(3)^{5-2}=\frac{135}{2}$
View full question & answer→MCQ 1792 Marks
$\lim _{x \rightarrow 2} \frac{x^{100}-2^{100}}{x^{77}-2^{77}}$ is equal to
- A
$\frac{100}{77}$
- B
$\frac{100}{77}\left(2^{22}\right)$
- C
$\frac{100}{77}\left(2^{21}\right)$
- ✓
$\frac{100}{77}\left(2^{23}\right)$
AnswerCorrect option: D. $\frac{100}{77}\left(2^{23}\right)$
(D)
$\lim _{x \rightarrow a} \frac{x^m-a^m}{x^n-a^n}=\frac{m}{n} a^{m-n}$
$\lim _{x>2} \frac{x^{100}-2^{100}}{x^{77}-2^{77}}=\frac{100}{77}$$(2)^{100-77}=\frac{100}{77}\left(2^{23}\right)$
View full question & answer→MCQ 1802 Marks
$\lim _{x \rightarrow \infty} \cos x=$
Answer(D)
Since $\cos x$ can have any value from -1 to 1 .
∴ limit does not exist.
View full question & answer→MCQ 1812 Marks
$\lim _{x \rightarrow \infty} \frac{\sin x}{x}=$
Answer(B)
Let $x=\frac{1}{y}$ or $y=\frac{1}{x}$,
so that $x \rightarrow \infty, y \rightarrow 0$
$\therefore \quad \lim _{x \rightarrow \infty}\left(\frac{\sin x}{x}\right)=\lim _{y \rightarrow 0}\left(y \cdot \sin \frac{1}{y}\right)$
$=\lim _{y \rightarrow 0} y \times \lim _{y \rightarrow 0} \sin \frac{1}{y}=0$
View full question & answer→MCQ 1822 Marks
The value of $\lim _{n \rightarrow \infty} \frac{x^n}{x^n+1}$, where $x < -1$, is
- A
$\frac{1}{2}$
- B
$\frac{-1}{2}$
- ✓
- D
Answer(C)
$\lim _{n \rightarrow \infty} \frac{x^n}{x^n\left(1+\frac{1}{x^n}\right)}=\lim _{n \rightarrow \infty} \frac{1}{\left(1+\frac{1}{x^n}\right)}=1$
View full question & answer→MCQ 1832 Marks
$\lim _{x \rightarrow \infty}\left(\frac{\sqrt{x^2-1}}{2 x+1}\right)=$
- A
- ✓
$\frac{1}{2}$
- C
- D
$-\frac{1}{2}$
AnswerCorrect option: B. $\frac{1}{2}$
(B)
$\lim _{x \rightarrow \infty}\left(\frac{\sqrt{x^2-1}}{2 x+1}\right)=\lim _{x \rightarrow \infty} \frac{x \sqrt{1-\frac{1}{x^2}}}{x\left(2+\frac{1}{x}\right)}$
$=\frac{\sqrt{1-0}}{2+0}$
$=\frac{1}{2}$
View full question & answer→MCQ 1842 Marks
$\lim _{x \rightarrow \infty}\left(\frac{x^3}{3 x^2-4}-\frac{x^2}{3 x+2}\right)$ is equal to
- A
$\frac{-1}{4}$
- B
$\frac{-1}{2}$
- C
$0$
- ✓
$\frac{2}{9}$
AnswerCorrect option: D. $\frac{2}{9}$
(D)
$\lim _{x \rightarrow \infty}\left(\frac{x^3}{3 x^2-4}-\frac{x^2}{3 x+2}\right)$
$=\lim _{x \rightarrow \infty}\left[\frac{3 x^4+2 x^3-3 x^4+4 x^2}{\left(3 x^2-4\right)(3 x+2)}\right]$
$=\lim _{x \rightarrow \infty} \frac{2 x^2(x+2)}{\left(3 x^2-4\right)(3 x+2)}$
$=\lim _{x \rightarrow \infty} \frac{2\left(1+\frac{2}{x}\right)}{\left(3-\frac{4}{x^2}\right)\left(3+\frac{2}{x}\right)}=\frac{2}{9}$
View full question & answer→MCQ 1852 Marks
The value of $\lim _{x \rightarrow \infty} \frac{(x+1)(3 x+4)}{x^2(x-8)}$ is equal to
Answer(D)
$\lim _{x \rightarrow \infty} \frac{(x+1)(3 x+4)}{x^2(x-8)}=\lim _{x \rightarrow \infty}\left[\frac{x\left(1+\frac{1}{x}\right) x\left(3+\frac{4}{x}\right)}{x^3\left(1-\frac{8}{x}\right)}\right]$
$=\lim _{x \rightarrow \infty}\left[\frac{1}{x} \frac{\left(1+\frac{1}{x}\right)\left(3+\frac{4}{x}\right)}{\left(1-\frac{8}{x}\right)}\right]=0$
View full question & answer→MCQ 1862 Marks
$\lim _{x \rightarrow-\infty} \frac{(2 x-3)(3 x-4)}{(4 x-5)(5 x-6)}=$
- A
$0$
- B
$\frac{1}{10}$
- C
$\frac{1}{5}$
- ✓
$\frac{3}{10}$
AnswerCorrect option: D. $\frac{3}{10}$
(D)
$\lim _{x \rightarrow \infty} \frac{(2 x-3)(3 x-4)}{(4 x-5)(5 x-6)}=\lim _{x \rightarrow \infty} \frac{x\left(2-\frac{3}{x}\right) x\left(3-\frac{4}{x}\right)}{x\left(4-\frac{5}{x}\right) x\left(5-\frac{6}{x}\right)}$
$=\frac{(2-0)(3-0)}{(4-0)(5-0)}$
$=\frac{6}{20}=\frac{3}{10}$
View full question & answer→MCQ 1872 Marks
$\lim _{x \rightarrow-\infty} \frac{3 x^3+2 x^3-7 x+9}{4 x^3+9 x-2}$ is equal to
- A
$\frac{2}{9}$
- B
$\frac{1}{2}$
- C
$\frac{-9}{2}$
- ✓
$\frac{3}{4}$
AnswerCorrect option: D. $\frac{3}{4}$
(D)
$\lim _{x \rightarrow \infty} \frac{3 x^3+2 x^2-7 x+9}{4 x^3+9 x-2}=\lim _{x \rightarrow \infty} \frac{3+\frac{2}{x}-\frac{7}{x^2}+\frac{9}{x^3}}{4+\frac{9}{x^2}-\frac{2}{x^3}}$
$=\frac{3}{4}$
View full question & answer→MCQ 1882 Marks
$\lim _{x \rightarrow \infty}\left(\frac{a x^2+b x+c}{p x^3+q x+r}\right)$ is equal to $(a, p \neq 0)$
- A
$\frac{ a }{ p }$
- B
$\frac{c}{r}$
- ✓
$0$
- D
$\frac{b}{q}$
Answer(C)
Here, degree of $N ^{ r }<$ degree of $D ^{ r }$
$\therefore \quad \lim _{x \rightarrow \infty}\left(\frac{ a x^2+ b x+ c }{ p x^3+ q x+ r }\right)=0$
View full question & answer→MCQ 1892 Marks
$\lim _{x \rightarrow \infty} \frac{2 x^2+3 x+4}{3 x^2+3 x+4}=$
- A
- B
- ✓
$\frac{2}{3}$
- D
$\frac{3}{2}$
AnswerCorrect option: C. $\frac{2}{3}$
(C)
$\lim _{x \rightarrow \infty} \frac{2 x^2+3 x+4}{3 x^2+3 x+4}=\lim _{x \rightarrow \infty} \frac{2+\frac{3}{x}+\frac{4}{x^2}}{3+\frac{3}{x}+\frac{4}{x^2}}=\frac{2}{3}$
Alternate method:
$\lim _{x \rightarrow \infty}\left(\frac{a x ^ 2+b x+c}{d x ^ 2+e x+f}\right)=\frac{a}{d}$
Here, degree of $N ^{ r }=$ degree of $D ^{ r }$
$\lim _{x \rightarrow \infty} \frac{2 x^2+3 x+4}{3 x^2+3 x+4}=\frac{2}{3}$
View full question & answer→MCQ 1902 Marks
The value of $\lim _{x \rightarrow e} \frac{\log x-1}{x-e}$ is
AnswerCorrect option: B. $\frac{1}{ e }$
(B)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow e } \frac{\log x-1}{x- e }=\lim _{x \rightarrow e } \frac{1}{x}=\frac{1}{ e }$
View full question & answer→MCQ 1912 Marks
$\lim _{x \rightarrow 0}\left[\frac{\log 5+\log \frac{(x+1)}{5}}{x}\right]$ is equal to
- ✓
- B
- C
$-\frac{1}{2}$
- D
$\frac{1}{2}$
Answer(A)
$\lim _{x \rightarrow 0} \frac{\log 5+\log \frac{(x+1)}{5}}{x}=\lim _{x \rightarrow 0} \frac{\log \left(5 \times \frac{x+1}{5}\right)}{x}=1$
View full question & answer→MCQ 1922 Marks
$\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{3^x-1}=$
- A
$\log _e 3$
- B
$0$
- C
- ✓
$\log _3 e$
AnswerCorrect option: D. $\log _3 e$
(D)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{3^x-1}=\lim _{x \rightarrow 0} \frac{\frac{1}{1+x}}{3^x \log _e 3}=\frac{1}{\log _e 3}=\log _3 e$
Alternate method:
$\lim _{x \rightarrow 0} \frac{\frac{\log _e(1+x)}{x}}{\frac{3^x-1}{x}}=\frac{\log e }{\log 3}=\log _3 e$
View full question & answer→MCQ 1932 Marks
$\lim _{x \rightarrow 0} \frac{\tan x}{\log (1+x)}$ is equal to
Answer(A)
$\lim _{x \rightarrow 0} \frac{\tan x}{\log (1+x)}=\lim _{x \rightarrow 0} \frac{\frac{\tan x}{x}}{\frac{\log (1+x)}{x}}=1$
View full question & answer→MCQ 1942 Marks
$\lim _{x \rightarrow 0} \frac{\log \left(1+\frac{8 x}{3}\right)}{x}$ is equal to
- ✓
$\frac{8}{3}$
- B
$\frac{-8}{3}$
- C
$\frac{3}{8}$
- D
$\frac{-3}{8}$
AnswerCorrect option: A. $\frac{8}{3}$
(A)
$\lim _{x \rightarrow 0} \frac{\log \left(1+\frac{8 x}{3}\right)}{x}=\lim _{x \rightarrow 0} \frac{\log \left(1+\frac{8 x}{3}\right)}{\frac{8}{3} x} \cdot \frac{8}{3}$
$=\frac{8}{3}$
View full question & answer→MCQ 1952 Marks
$\lim _{x \rightarrow 1} \frac{\log x}{x-1}=$
Answer(A)
$\lim _{x \rightarrow 1} \frac{\log [(x-1)+1]}{x-1}=1$
Alternate method:
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 1} \frac{\log x}{x-1}=\lim _{x \rightarrow 1} \frac{\frac{1}{x}}{1}=1$
View full question & answer→MCQ 1962 Marks
The value of $\lim _{x \rightarrow 0} \frac{2}{x} \log (1+x)$ is equal to
Answer(D)
$\lim _{x \rightarrow 0} \frac{2}{x} \log (1+x)=2 \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}$
$=2(1)=2$
View full question & answer→MCQ 1972 Marks
$\lim _{x \rightarrow 0}(1-a x)^{\frac{1}{x}}=$
AnswerCorrect option: B. $e ^-a$
(B)
$\lim _{x \rightarrow 0}(1- a x)^{\frac{1}{x}}=\lim _{x \rightarrow 0}\left\{[1+(- a x)]^{\frac{1}{-\frac{1}{2 x}}}\right\}^{-8}$
$= e ^{- a } \quad \ldots .\left[\because \lim (1+x)^{\frac{1}{x}}= e \right]$
View full question & answer→MCQ 1982 Marks
$\lim _{x \rightarrow 0} \frac{ a ^x- b ^x}{ e ^x-1}=$
AnswerCorrect option: A. $\log \left(\frac{a}{b}\right)$
(A)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{a^x-b^x}{e^x-1}=\lim _{x \rightarrow 0} \frac{a^x \log a-b^x \log b}{e^x}$
$=\log a-\log b=\log \left(\frac{a}{b}\right)$
View full question & answer→MCQ 1992 Marks
$\lim _{x \rightarrow 0} \frac{\left(1-e^x\right) \sin x}{x^2+x^3}$ is equal to
Answer(A)
$\lim _{x \rightarrow 0} \frac{\left(1- e ^x\right) \sin x}{x^2+x^3}$
$=-\lim _{x \rightarrow 0}\left(\frac{ e ^x-1}{x}\right) \times\left(\frac{\sin x}{x}\right) \times \frac{1}{1+x}$
$=-1 \times 1 \times 1=-1$
View full question & answer→MCQ 2002 Marks
$\lim _{x \rightarrow 0} \frac{e^{a x}-e^{b x}}{x}=$
- A
$\alpha+\beta$
- B
$\frac{1}{\alpha}+\beta$
- C
$\alpha^2+\beta^2$
- ✓
$\alpha-\beta$
AnswerCorrect option: D. $\alpha-\beta$
(D)
$\lim _{x \rightarrow 0} \frac{ e ^{a x}- e ^{\beta x}}{x}=\lim _{x \rightarrow 0} \frac{ e ^{a x}-1- e ^{\beta x}+1}{x}$
$=\alpha \lim _{x \rightarrow 0} \frac{ e ^{\alpha x}-1}{\alpha x}-\beta \lim _{x \rightarrow 0} \frac{ e ^{\beta x}-1}{\beta x}$
$=\alpha-\beta$
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