Limits and Derivatives

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New answer posted•a year ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$l i m_{x \to a} \frac{s i n x - s i n a}{\sqrt[]{x} - \sqrt[]{a}}$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to a}{l i m} \frac{s i n x - s i n a}{\sqrt[]{x} - \sqrt[]{a}} \\ = \underset{x \to a}{l i m} \frac{s i n x - s i n a}{\sqrt[]{x} - \sqrt[]{a}} \times \frac{\sqrt[]{x} + \sqrt[]{a}}{\sqrt[]{x} + \sqrt[]{a}} = \underset{x \to a}{l i m} \frac{(s i n x - s i n a) (\sqrt[]{x} + \sqrt[]{a})}{x - a} \\ = \underset{x \to a}{l i m} \frac{(2 c o s \frac{x + a}{2} . s i n \frac{x - a}{2}) (\sqrt[]{x} + \sqrt[]{a})}{x - a} [? s i n A - s i n B = 2 c o s \frac{A + B}{2} . s i n \frac{A - B}{2}] \\ = \underset{\frac{x - a}{2} \to 0}{l i m} (2 c o s \frac{x + a}{2} . \frac{s i n \frac{x - a}{2}}{2 \times \frac{x - a}{2}}) (\sqrt[]{x} + \sqrt[]{a}) \\ = \underset{x \to a}{l i m} c o s (\frac{x + a}{2}) (\sqrt[]{x} + \sqrt[]{a}) [? \underset{\frac{x - a}{2} \to 0}{l i m} \frac{s i n \frac{x - a}{2}}{\frac{x - a}{2}} = 1] \\ T a k i n g, w e h a v e \\ = c o s (\frac{a + a}{2}) (\sqrt[]{a} + \sqrt[]{a}) = c o s a \times 2 \sqrt[]{a} = 2 \sqrt[]{a} . c o s a \end{array}$

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New answer posted•a year ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$\underset{x \to 0}{l i m} \frac{s i n 2 x + 3 x}{2 x + t a n 3 x}$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n 2 x + 3 x}{2 x + t a n 3 x} \\ = \underset{x \to 0}{l i m} \frac{(\frac{s i n 2 x + 3 x}{2 x} \times 2 x)}{(\frac{2 x + t a n 3 x}{3 x} \times 3 x)} = \underset{x \to 0}{l i m} \frac{(\frac{s i n 2 x}{2 x} + \frac{3 x}{2 x}) \times 2 x}{(\frac{2 x}{3 x} + \frac{t a n 3 x}{3 x}) \times 3 x} \\ = \frac{(\underset{2 x \to 0}{l i m} \frac{s i n 2 x}{2 x} + \frac{3}{2})}{[\frac{2}{3} + \underset{3 x \to 0}{l i m} \frac{t a n 3 x}{3 x}]} \times \frac{2}{3} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ = (\frac{1 + \frac{3}{2}}{\frac{2}{3} + 1}) \times \frac{2}{3} [? \underset{x \to 0}{l i m} \frac{t a n x}{x} = 1] \\ = \frac{5 / 2}{5 / 3} \times \frac{2}{3} = \frac{3}{2} \times \frac{2}{3} = 1 \end{array}$

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New answer posted•a year ago

Limits and Derivatives Exemplar Solution

Kindly consider the following
$l i m_{x \to \frac{π}{6}} \frac{\sqrt[]{3} s i n x - c o s x}{x - \frac{π}{6}}$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{6}}{l i m} \frac{\sqrt[]{3} s i n x - c o s x}{x - \frac{π}{6}} \\ = \underset{x \to \frac{π}{6}}{l i m} \frac{2 [\frac{\sqrt[]{3}}{2} s i n x - \frac{1}{2} c o s x]}{x - \frac{π}{6}} = \underset{x \to \frac{π}{6}}{l i m} \frac{2 [c o s \frac{π}{6} s i n x - s i n \frac{π}{6} c o s x]}{x - \frac{π}{6}} \\ = \underset{x \to \frac{π}{6}}{l i m} \frac{2 s i n (x - \frac{π}{6})}{x - \frac{π}{6}} [? s i n (A - B) = s i n A c o s B - c o s A s i n B] \\ = 2 \underset{x \to \frac{π}{6}}{l i m} \frac{s i n (x - \frac{π}{6})}{x - \frac{π}{6}} = 2.1 = 2 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

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New answer posted•a year ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$l i m_{x \to \frac{π}{4}} \frac{s i n x - c o s x}{x - \frac{π}{4}}$

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{4}}{l i m} \frac{s i n x - c o s x}{x - \frac{π}{4}} \\ = \underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt[]{2} (\frac{1}{\sqrt[]{2}} s i n x - \frac{1}{\sqrt[]{2}} c o s x)}{x - \frac{π}{4}} = \underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt[]{2} (c o s \frac{π}{4} s i n x - s i n \frac{π}{4} c o s x)}{x - \frac{π}{4}} \\ = \underset{x \to \frac{π}{4}}{l i m} \frac{\sqrt[]{2} s i n (x - \frac{π}{4})}{x - \frac{π}{4}} [? s i n (a - b) = s i n a c o s b - c o s a s i n b] \\ = \sqrt[]{2} \underset{x \to \frac{π}{4}}{l i m} \frac{s i n (x - \frac{π}{4})}{x - \frac{π}{4}} = \sqrt[]{2} . 1 = \sqrt[]{2} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

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New answer posted•a year ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$\underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{1 - c o s 6 x}}{\sqrt[]{2} (\frac{π}{3} - x)}$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{1 - c o s 6 x}}{\sqrt[]{2} (\frac{π}{3} - x)} \\ = \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{2 s i n^{2} 3 x}}{\sqrt[]{2} (\frac{π}{3} - x)} [? 1 - c o s θ = 2 s i n^{2} \frac{θ}{2}] \\ = \underset{x \to \frac{π}{3}}{l i m} \frac{\sqrt[]{2} s i n 3 x}{\sqrt[]{2} (\frac{π}{3} - x)} = \underset{π - 3 x \to 0}{l i m} \frac{3 . s i n (π - 3 x)}{π - 3 x} \\ = 3 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

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New answer posted•a year ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 0} \frac{1 - c o s m x}{1 - c o s n x}$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{1 - c o s m x}{1 - c o s n x} \\ = \underset{x \to 0}{l i m} (\frac{1 - 1 + 2 s i n^{2} \frac{m x}{2}}{1 - 1 + 2 s i n^{2} \frac{n x}{2}}) [? c o s m x = 1 - 2 s i n^{2} \frac{m x}{2}] \\ = \underset{x \to 0}{l i m} (\frac{2 s i n^{2} \frac{m x}{2}}{2 s i n^{2} \frac{n x}{2}}) = \underset{x \to 0}{l i m} {(\frac{s i n \frac{m x}{2}}{s i n \frac{n x}{2}})}^{2} = \frac{\underset{x \to 0}{l i m} (\frac{s i n \frac{m x}{2}}{\frac{m x}{2}} \times \frac{m x}{2})}{\underset{x \to 0}{l i m} (\frac{s i n \frac{n x}{2}}{\frac{n x}{2}} \times \frac{n x}{2})} \\ = \frac{1 . \frac{m^{2} x^{2}}{4}}{1 . \frac{n^{2} x^{2}}{4}} = \frac{m^{2}}{n^{2}} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

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New answer posted•a year ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 0} \frac{2 s i n x - s i n 2 x}{x^{3}}$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{2 s i n x - s i n 2 x}{x^{3}} \\ = \underset{x \to 0}{l i m} \frac{2 s i n x - 2 s i n x c o s x}{x^{3}} = \underset{x \to 0}{l i m} \frac{2 s i n x (1 - c o s x)}{x^{3}} \\ = \underset{x \to 0}{l i m} \frac{2 s i n x}{x} (\frac{1 - c o s x}{x^{2}}) = \underset{x \to 0}{l i m} 2 (\frac{s i n x}{x}) (\frac{2 s i n^{2} x / 2}{x^{2}}) \\ = \underset{x \to 0}{l i m} 2 (\frac{s i n x}{x}) (2 \frac{s i n^{2} x / 2}{\frac{x^{2}}{4}} \times \frac{1}{4}) = \underset{x \to 0}{l i m} 2 (\frac{s i n x}{x}) 2 {(\frac{s i n x / 2}{\frac{x}{2}})}^{2} \times \frac{1}{4} \\ = \underset{x \to 0}{l i m} \frac{4}{4} (\frac{s i n x}{x}) \underset{\frac{x}{2} \to 0}{l i m} {(\frac{s i n x / 2}{\frac{x}{2}})}^{2} = 1.1 . {(1)}^{2} = 1 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

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New answer posted•a year ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following
$l i m_{x \to 0} \frac{1 - c o s 2 x}{x^{2}}$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{1 - c o s 2 x}{x^{2}} \\ = \underset{x \to 0}{l i m} \frac{2 s i n^{2} x}{x^{2}} [c o s 2 x = 1 - 2 s i n^{2} x] \\ = \underset{x \to 0}{l i m} 2 {(\frac{s i n x}{x})}^{2} = 2 \times 1 = 2 [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

Upvote

New answer posted•a year ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to 0} \frac{s i n^{2} 2 x}{s i n^{2} 4 x}$

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n^{2} 2 x}{s i n^{2} 4 x} \\ = \underset{x \to 0}{l i m} \frac{s i n^{2} 2 x}{s i n^{2} 2 (2 x)} \\ = \underset{x \to 0}{l i m} \frac{s i n^{2} 2 x}{4 s i n^{2} 2 x . c o s^{2} 2 x} [s i n 2 x = 2 s i n x c o s x] \\ = \underset{x \to 0}{l i m} \frac{1}{4 c o s^{2} 2 x} \\ T a k i n g limit, w e h a v e \\ = \frac{1}{4 . c o s^{2} 0} = \frac{1}{4} \end{array}$

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New answer posted•a year ago

Limits and Derivatives Maths NCERT Exemplar Solutions Class 11th Chapter Thirteen

Kindly consider the following

$l i m_{x \to a} \frac{s i n 3 x}{s i n 7 x}$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t \underset{x \to 0}{l i m} \frac{s i n 3 x}{s i n 7 x} \\ = \underset{x \to 0}{l i m} \frac{\frac{s i n 3 x}{3 x} \times 3 x}{\frac{s i n 7 x}{7 x} \times 7 x} = \frac{\underset{3 x \to 0}{l i m} (\frac{s i n 3 x}{3 x})}{\underset{3 x \to 0}{l i m} (\frac{s i n 7 x}{7 x})} \times \frac{3}{7} \\ = \frac{1}{1} \times \frac{3}{7} = \frac{3}{7} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \end{array}$

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