MCQ 11 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \lim\limits_{\text{x}\rightarrow 0}\frac{3^\text{x}-2^\text{x}}{\tan\text{x}}$ is equal to $\log\big(\frac{3}{2}\big)$
Reason $(R) \lim\limits_{\text{x}\rightarrow 0} \frac{\log(1+\text{x)}}{\tan\text{x}}$ is equal to $2.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- ✓
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: C. $A$ is true; $R$ is false
Assertion Given limit $=\lim\limits_{\text{x}\rightarrow0}\frac{3^\text{x}-2^\text{x}}{\tan\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{(3^\text{x}-1)-(2^\text{x}-1)}{\text{x}}\times\frac{\text{x}}{\tan\text{x}}$
$=\Big(\lim\limits_{\text{x}\rightarrow0}\frac{3^\text{x}-1}{\text{x}}-\lim\limits_{\text{x}\rightarrow0}\frac{2^\text{x}-1}{\text{x}}\Big)\times\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}}{\tan\text{x}}$
$=(\log3-\log2)\times1$
$=\log\Big(\frac{3}{2}\Big)$
Reason $=\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\tan\text{x}}$
$\lim\limits_{\text{x}\rightarrow0}\frac{\log(1+\text{x})}{\text{x}}\times\frac{\text{x}}{\tan\text{x}}$
$=1\times1=1 $
View full question & answer→MCQ 21 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)\ \lim\limits_{\text{x}\rightarrow\pi} \frac{\sin(\pi-\text{x)}}{\pi(\pi-\text{x})}$ is equal to $\pi$
Reason $(R)\ \lim\limits_{\text{x}\rightarrow 0} \frac{\cos\text{x}}{\pi-\text{x}}$ is equal to $\frac{1}{\pi}.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- ✓
$A$ is false; $R$ is true.
AnswerCorrect option: D. $A$ is false; $R$ is true.
Assertion Given, $\lim\limits_\text{x}\rightarrow\pi\frac{\sin(\pi-\text{x)}}{\pi(\pi-\text{x)}}$
Let $\pi - \text{x} =\text{h},$ As $\text{x} \rightarrow \pi,$ then $\text{ h} \rightarrow 0 $
$\therefore\lim\limits_{\text{x}\rightarrow\pi}\frac{\sin(\pi-\text{x})}{\pi(\pi-{\text{x})}\lim\limits_{\text{h}\rightarrow0}}\frac{\sin\text{h}}{\pi\text{h}}$
$\lim\limits_{\text{h}\rightarrow0}\frac{1}{\pi}\Big[\because\lim\limits_{\text{h}\rightarrow0}\frac{\sin\text{h}}{\text{h}}=1\Big]$
Reason Given, $=\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{x}}{\pi-\text{x}}$
Put the limit directly, we get
$\frac{\cos0}{\pi-0}=\frac{1}{\pi}$
View full question & answer→MCQ 31 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)\ \lim\limits_{\text{x}\rightarrow0} \frac{\sin\text{ax}}{\sin\text{bx}}$ is equal to $\frac{\text{a}}{\text{b}}.$
Reason $(R)\ \lim\limits_{\text{x}\rightarrow0} \frac{\sin\text{x}}{\text{x}}=1.$
- ✓
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: A. $A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
Assertion Given, $\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}}{\sin\text{bx}}$
Multiplying and dividing by $(ax)$ and $(x),$ we get
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}}{\text{ax}}\times\frac{\text{bx}}{\sin\text{bx}}\times\frac{\text{ax}}{\text{bx}}$
$=1\times1\times\frac{\text{a}}{\text{b}}$
$=\frac{\text{a}}{\text{b}}$
$\Big[\because\lim\limits_{\text{a}\rightarrow0}\frac{\sin\text{ax}}{\text{ax}}=\lim\limits_{\text{x}\rightarrow0}\frac{\text{bx}}{\sin\text{bx}}=1\Big]$
View full question & answer→MCQ 41 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A)\ \lim\limits_{\text{x}\rightarrow0} \frac{\ell^{3+\text{x}}-\sin\text{x-}\ell^3}{\text{x}}$ is equal to $\ell^3+1.$
Reason $(R)\ \lim\limits_{\text{x}\rightarrow0} \frac{\tan4\text{x}}{\sin2\text{x}}$ is equal to $2.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- ✓
$A$ is false; $R$ is true.
AnswerCorrect option: D. $A$ is false; $R$ is true.
Assertion Given, limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\ell^{3+\text{x}}-\sin\text{x}-\ell^3}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\Big[\frac{\ell^3\text{(e}^\text{x}-1)}{\text{x}}-\frac{\sin\text{x}}{\text{x}}\Big]=\ell^3-1$
Reason Given limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\tan4\text{x}}{\sin2\text{x}}$
$\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\tan4\text{x}}{4\text{x}}\Big)\Big|\Big|\Big(\frac{2\text{x}}{\sin2\text{x}}\Big)\Big|\Big|\Big(\frac{4\text{x}}{2\text{x}}\Big)$
$\lim\limits_{4\text{x}\rightarrow0}\frac{\tan4\text{x}}{4\text{x}}\times\frac{1}{\lim\limits_{2\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}}\times\frac{4\text{x}}{2\text{x}}$
$=1\times1\times2=2 $
View full question & answer→MCQ 51 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{x}\cos\text{x}}{\text{b}\sin\text{x}}$ is equal to $\frac{\text{a+1}}{\text{b}}.$
Reason $(R) \lim\limits_{\text{x}\rightarrow0}\text{x}\sec\text{x}$ is equal to $1.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- ✓
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: C. $A$ is true; $R$ is false
Assertion Given, $\lim\limits_{\text{x}\rightarrow0}\frac{\text{ax}+\text{x}\cos\text{x}}{\text{b}\sin\text{x}}$
Dividing each term by $x,$ we get
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\text{ax}}{\text{ax}}+\frac{\text{x}\cos\text{x}}{\text{x}}}{\frac{\text{b}\sin\text{x}}{\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0} \frac{\text{a}+\cos\text{x}}{\text{b}\Big|\frac{\sin\text{x}}{\text{x}}\Big|}$
$=\frac{\text{a}+\cos0}{\text{b}\times1}$
$=\frac{\text{a}+1}{\text{b}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
Reason $=\lim\limits_{\text{x}\rightarrow0}\text{x}\text{ sec}\text{ x}=0 \times\sec0=0 \times1=0$
View full question & answer→MCQ 61 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A): \lim\limits_{\text{x}\rightarrow 0} \frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}$ is equal to $-2.$
Reason $(R): \lim\limits_{\text{x}\rightarrow 1} $ $(5x^3 +5x +1)$ is equal to $11.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- ✓
$A$ is false; $R$ is true.
AnswerCorrect option: D. $A$ is false; $R$ is true.
Assertion $\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}+\text{bx}}{\text{ax}+\sin\text{bx}}$
Dividing each term by $x,$ we get
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\sin\text{ax}}{\text{x}}+\frac{\text{bx}}{\text{x}}}{\frac{\text{ax}}{\text{x}}+\frac{\sin\text{bx}}{\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\text{a}\sin\text{ax}}{\text{ax}}+\text{b}}{\text{a}+\frac{\text{b}\sin\text{bx}}{\text{bx}}}$
$=\frac{\text{a}\times1+\text{b}}{\text{a}+\text{b}\times1}=\frac{\text{a}+\text{b}}{\text{a}+\text{b}}=1$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
Reason $\lim\limits_{\text{x}\rightarrow1}\ (5\text{x} + 5\text{x} +1) $
$5(1)^\circ + 5(1)41=54+541=11$
View full question & answer→MCQ 71 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \lim\limits_{\text{x}\rightarrow 0} \frac{\text{x}-1}{\log_\ell\text{x}}$ is equal to $1.$
Reason $(R) \lim\limits_{\text{x}\rightarrow 0}\frac{\log(\sin\text{x+1)}}{\text{x}}$ is equal to $0$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- ✓
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: C. $A$ is true; $R$ is false
Assertion Given limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}-1}{\log_\text{e}\text{x}}$
$\text{put } \text{x}=1+\text{h}$ as $\text{ x}\rightarrow1,\text{h}\rightarrow0 $
$\therefore\lim\limits_{\text{h}\rightarrow0}\frac{1+\text{h}-1}{\log_\text{e}(1+\text{h)}}$
$=\frac{1}{\lim\limits_{\text{h}\rightarrow0}\frac{\log(1+\text{h)}}{\text{h}}}$
$=1$
Reason Given, limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\log(\sin\text{x+1)}}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\log(\sin\text{x+1)}}{\sin\text{x}}\times\frac{\sin\text{x}}{\text{x}}$
$=1$
View full question & answer→MCQ 81 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \lim\limits_{\text{x}\rightarrow 0} \frac{\sin\text{ax}}{\text{bx}}$ is equal to $\frac{\text{a}}{\text{b}}.$
Reason $(R) \lim\limits_{\text{x} \rightarrow 0} \frac{\sin\text{x}}{\text{x}}=1.$
- ✓
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: A. $A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
Assertion Given,
$\lim\limits_{\text{a}\rightarrow0}\frac{\sin\text{ax}}{\text{bx}}=\lim\limits_{\text{x}\rightarrow0}\frac{(\text{a})\sin\text{ax}}{\text{b}(\text{bx})}$
$[$dividing and multiplying by $a]$
$=\frac{\text{a}}{\text{b}}\times1$
$=\frac{\text{a}}{\text{b}}\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{ax}}{\text{ax}}=1\Big]$
View full question & answer→MCQ 91 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \lim\limits_{\text{x}\rightarrow 0}\frac{\ell^\text{x}-1}{\sqrt{1-\cos2\text{x}}}$ — is equal to $\frac{1}{\sqrt{2}}$
Reason $(R) \lim\limits_{\text{x}\rightarrow 1} \frac{\text{x}^2-1}{\text{x}-1}$ is equal to $1.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- ✓
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: C. $A$ is true; $R$ is false
Assertion Given, limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\sqrt{1-\cos2\text{x}}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\sqrt{1-\cos2\text{x}}}$
$=\frac{1}{\sqrt{2}}\lim\limits_{\text{x}\rightarrow0}\frac{\text{e}^\text{x}-1}{\text{x}}.\frac{\text{x}}{\sin\text{x}}$
$=\frac{1}{\sqrt{2}}\times1\times=\frac{1}{\sqrt{2}}$
Reason Given, limit $=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}^2-1)}{(\text{x}-1)}$
$=\lim\limits_{\text{x}\rightarrow1}\frac{(\text{x}-1)\text{(x+1)}}{(\text{x}-1)}$
$=\lim\limits_{\text{x}\rightarrow1}(\text{x}+1)$
$=2$
View full question & answer→MCQ 101 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \lim\limits_{\text{x} \rightarrow 1}\frac{\text{ax}^2+\text{bx+}\text{c}}{\text{cx}^2+\text{bx}+\text{a}}$ equal to $1,$ where $\text{a}+\text{b}+\text{c}\neq0.$
Reason $(R) \lim\limits_{\text{x} \rightarrow -2}\frac{\frac{1}{\text{x}}+\frac{1}{2}}{\text{x}+2}$ is equal to $\frac{1}{4}.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- ✓
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: C. $A$ is true; $R$ is false
Assertion Given,
$\lim\limits_{\text{x} \rightarrow 1}\frac{\text{ax}^2+\text{bx+}\text{c}}{\text{cx}^2+\text{bx}+\text{a}}$
$=\frac{\text{a}\times(1)^2+\text{b}\times1+\text{c}}{\text{c}\times(1)^2+\text{b}\times1+\text{a}}$
$=\frac{\text{a}+\text{b}+\text{c}}{\text{c}+\text{b}+\text{c}}=1$
Reason $\lim\limits_{\text{x} \rightarrow -2}\frac{\frac{1}{\text{x}}+\frac{1}{2}}{\text{x}+2}$
$=\lim\limits_{\text{x}\rightarrow-2}\frac{(2+\text{x)}}{2\text{x}(\text{x}+2)}$
$=\lim\limits_{\text{x}\rightarrow-2}\frac{1}{2\text{x}}$
$=\frac{1}{2(-2)}$
$=-\frac{1}{4}$
View full question & answer→MCQ 111 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \lim\limits_{\text{x}\rightarrow 0} \frac{3{^{2+\text{x}}-9}}{\text{x}}$ is equal to $9\log2$
Reason $(R) \lim\limits_{\text{x}\rightarrow 0}\frac{\text{a}^{\sin\text{x}}-1}{\sin\text{x}}$ is equal to $\log\text{a}.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- ✓
$A$ is false; $R$ is true.
AnswerCorrect option: D. $A$ is false; $R$ is true.
Assertion Given, limit $=\lim\limits_{\text{x}\rightarrow0}\frac{3^{2+\text{x}}-9}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{3^2(3^\text{x}-1)}{\text{x}}=9\log3$
Reason Given, limit $=\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^{\sin\text{x}}-1}{\sin\text{x}}$
Let $\text{ y}=\sin\text{x}$
then,$\text{ y}\rightarrow0\text{ as}\text{ x}\rightarrow0$
$\therefore\lim\limits_{\text{x}\rightarrow0}\frac{\text{a}^{\sin\text{x}}-1}{\sin\text{x}}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{\text{a}^\text{y}-1}{\text{y}}=\log\text{a}$
View full question & answer→MCQ 121 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \lim\limits_{\text{x}\rightarrow 0} \frac{\ell^\text{x}-\ell^\text{-x}}{\text{x}}$ is equal to $2$
Reason $(R) \lim\limits_{\text{x}\rightarrow0} \frac{\ell^\text{x}-1}{\text{x}}=1.$
- ✓
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: A. $A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
Assertion Given, limit $=\lim\limits_\text{x}\rightarrow\frac{\ell^\text{x}-\ell^-{\text{x}}}{\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\ell^{2\text{x}}-1}{\text{x}\ell^\text{x}}$
$=\lim\limits_{2\text{x}\rightarrow0}\frac{\ell^{2\text{x}}-1}{2\text{x}}\times\lim\limits_{2\text{x}\rightarrow0}\frac{2}{\ell^\text{x}}$
$=1\times\frac{2}{1}=2$
View full question & answer→MCQ 131 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \lim\limits_{\text{x}\rightarrow0} \frac{\ell^{\tan\text{x}}-1}{\text{x}}$ is equal to $1.$
Reason $(R) \lim\limits_{\text{x}\rightarrow 0} \Big(\frac{\ell^{4\text{x}}-1}{\text{x}}\Big)$ is equal to $2.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- ✓
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: C. $A$ is true; $R$ is false
Assertion Given limit
$=\lim\limits_{\text{x}\rightarrow0}\frac{\ell^{\tan\text{x}}-1}{\tan\text{x}}\times\frac{\tan\text{x}}{\text{x}}$
$=1\times1=1$
Reason Given limit
$\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\ell^{4\text{x}}-1}{\text{x}}\Big)$
$\lim\limits_{4\text{x}\rightarrow0}4\Big(\frac{\ell^{4\text{x}}-1}{\text{x}}\Big)=4$
View full question & answer→MCQ 141 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \lim\limits_{\text{x}\rightarrow 0} \frac{\ell^\text{x}-\ell^{\sin\text{x}}}{\text{x}-\sin\text{x}}$ is equal to $-1.$
Reason $(R) \lim\limits_{\text{x}\rightarrow 0}\Big(\frac{3^{2\text{x}}-2^{3\text{x}}}{\text{x}}\Big)$ is equal to $\log\big(\frac{9}{8}\big)$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- B
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- ✓
$A$ is false; $R$ is true.
AnswerCorrect option: D. $A$ is false; $R$ is true.
Assertion Given limit $=\lim\limits_{\text{a}\rightarrow0}\Big(\frac{\ell^\text{x}-\ell^{\sin\text{x}}}{\text{x}-\sin\text{x}}\Big)$
$=\lim\limits_{\text{a}\rightarrow0}\ell^{\sin\text{x}}\Big(\frac{\ell^{\text{x}-\sin\text{x}}-1}{\text{x}-\sin\text{x}}\Big)=\ell^{\sin0}\times1=1$
Reason Given limit $=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{3^{2\text{x}}-2^{3\text{x}}}{\text{x}}\Big)$
$=\lim\limits_{\text{x}\rightarrow0}\Big(\frac{(3^{2\text{x}}-1)-(2^{3\text{x}}-1}{\text{x}}\Big)$
$=2\times\lim\limits_{\text{x}\rightarrow0}\Big(\frac{3^{2\text{x}}-1}{2\text{x}}\Big)-3\times\lim\limits_{\text{x}\rightarrow0}\Big(\frac{2^{3\text{x}}-1}{3\text{x}}\Big)$
$= 2\log 3-3\log 2 $
$=\log 9-\log8 $
$=\log\Big(\frac{9}{8}\Big)$
View full question & answer→MCQ 151 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason$(s) (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion $(A) \lim\limits_{\text{x}\rightarrow0} \frac{\cos2\text{x-1}}{\cos\text{x}-1}$ is equal to $4$
Reason $(R) \lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1.$
- A
$A$ is true, $R$ is true; $R$ is acorrect explanation of $A.$
- ✓
$A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
- C
$A$ is true; $R$ is false
- D
$A$ is false; $R$ is true.
AnswerCorrect option: B. $A$ is true, $R$ is true; $R$ is not a correct explanation of $A.$
Assertion Given,
$\lim\limits_{\text{x}\rightarrow0}\Big(\frac{\sin\text{x}}{\text{x}}\Big)^2\times\Big(\frac{\frac{\text{x}}{2}}{\sin\frac{\text{x}}{2}}\Big)\times4$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\cos2\text{x}-1}{\cos\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{1-\cos2\text{x}}{1-\cos\text{x00}}=\lim\limits_{\text{x}\rightarrow0}\frac{2\sin^2\text{x}}{2\sin^2\frac{\text{x}}{2}}$
$\Big[\because1-\cos2\text{x}=2\sin^2\text{x}\text{ and }1-\cos\text{x}=2\sin^2\frac{\text{x}}{2}\Big]$
Multiplying and dividing by $x^2$ and then multiplying by $\frac{4}{4}$ in the numerater. we get
$=\lim\limits_{\text{x}\rightarrow0}\frac{\sin^2\text{x}}{\text{x}^2}\times\frac{4\times\frac{\text{x}^2}{4}}{\sin^2\frac{\text{x}}{2}}$
$=1\times1\times4$
$=4 $
View full question & answer→