\lim_{x \to b}\frac{\sqrt{x-a}-\sqrt{b-a}}{x^{2}-b^{2}}

  • Option 1)

    \frac{1}{4b\sqrt{a-b}}

  • Option 2)

    \frac{1}{4b\sqrt{b-a}}

  • Option 3)

    \frac{1}{4a\sqrt{a-b}}

  • Option 4)

    \frac{1}{b\sqrt{b-a}}

 

Answers (1)
A Aadil Khan

As we learnt in

Method of Rationalisation -

Rationalisation method is used when we have RADICAL SIGNS in an expression.(like  1/2,  1/3 etc) and there exists a negative sign between two terms of an algebraic expression.

- wherein

\lim_{x\rightarrow a}\:\frac{x-a}{\sqrt{x}-\sqrt{a}}


\therefore \:\frac{(x-a)(\sqrt{x}+\sqrt{a})}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}


=\sqrt{x}+\sqrt{a}

=\sqrt{a}+\sqrt{a}

=2\sqrt{a}

 

 

 \lim_{x\rightarrow b}\frac{\sqrt{x-a}-\sqrt{b-a}}{x^{2}-b^{2}}

\Rightarrow \lim_{x\rightarrow b}\frac{\sqrt{x-a}-\sqrt{b-a}}{x^{2}-b^{2}} \times \frac{\sqrt{x-a}+\sqrt{b-a}}{\sqrt{x-a}+\sqrt{b-a}}

= \lim_{x\rightarrow b} \frac{(x-a)-(b-a)}{(x^{2}-b^{2})\left(\sqrt{x-a}+\sqrt{b-a} \right )}

= \lim_{x\rightarrow b} \frac{(x-b)}{(x^{2}-b^{2})\left(\sqrt{x-a}+\sqrt{b-a} \right )}

= \lim_{x\rightarrow b} \frac{1}{(x+b)\left(\sqrt{x-a}+\sqrt{b-a} \right )}

= \lim_{x\rightarrow b} \frac{1}{2b \times 2 \sqrt{b-a}}=\frac{1}{4b \sqrt{b-a}}

 

 

 

 

 


Option 1)

\frac{1}{4b\sqrt{a-b}}

Incorrect

Option 2)

\frac{1}{4b\sqrt{b-a}}

Correct

Option 3)

\frac{1}{4a\sqrt{a-b}}

Incorrect

Option 4)

\frac{1}{b\sqrt{b-a}}

Incorrect

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