Question

Find the value of following limit:
${{\lim }_{x\to 0}}\dfrac{x\sqrt{{{y}^{2}}-{{(y-x)}^{2}}}}{{{\left( \sqrt{(8xy-4{{x}^{2}})}+\sqrt{8xy} \right)}^{3}}}$
(a) $\dfrac{1}{4y}$
(b) $\dfrac{1}{2}$
(c) $\dfrac{1}{2\sqrt{2}}$
(d) $\dfrac{1}{128y}$

Answer 1

Hint: We will first simplify the numerator and denominator by cancelling similar terms. Then we will take out the common factors of x from the numerator and denominator and cancel them out. Finally, we will substitute x equal to 0 to get our answer.

Complete step-by-step answer:
Before proceeding with the question we should understand the concept of limits. The concept of limits gives the value of an expression which is not the true value of it. The value obtained by limit is as close to the exact answer as possible. This becomes very helpful in expressions where the exact answer can never be known.
Now let ${{\lim }_{x\to 0}}\dfrac{x\sqrt{{{y}^{2}}-{{(y-x)}^{2}}}}{{{\left( \sqrt{(8xy-4{{x}^{2}})}+\sqrt{8xy} \right)}^{3}}}.........(1)$
Expanding and simplifying the numerator and denominator in equation (1) we get,
${{\lim }_{x\to 0}}\dfrac{x\sqrt{{{y}^{2}}-{{y}^{2}}-{{x}^{2}}+2xy}}{{{\left( \sqrt{(8xy-4{{x}^{2}})}+\sqrt{8xy} \right)}^{3}}}.........(2)$
Cancelling similar terms from equation (2) we get,
${{\lim }_{x\to 0}}\dfrac{x\sqrt{2xy-{{x}^{2}}}}{{{\left( \sqrt{(8xy-4{{x}^{2}})}+\sqrt{8xy} \right)}^{3}}}.........(3)$
Now taking out common terms from the numerator and denominator in equation (3) we get,

& {{\lim }_{x\to 0}}\dfrac{x\sqrt{x}\sqrt{2y-x}}{{{\left( \sqrt{x}\sqrt{(8y-4x)}+\sqrt{x}\sqrt{8y} \right)}^{3}}} \\\ & {{\lim }_{x\to 0}}\dfrac{x\sqrt{x}\sqrt{2y-x}}{{{\left( \sqrt{x} \right)}^{3}}{{\left( \sqrt{(8y-4x)}+\sqrt{8y} \right)}^{3}}} \\\ & {{\lim }_{x\to 0}}\dfrac{x\sqrt{x}\sqrt{2y-x}}{x\sqrt{x}{{\left( \sqrt{(8y-4x)}+\sqrt{8y} \right)}^{3}}}........(4) \\\ \end{aligned}$$ Now cancelling similar terms from the numerator and denominator in equation (4) we get, $${{\lim }_{x\to 0}}\dfrac{\sqrt{2y-x}}{{{\left( \sqrt{(8y-4x)}+\sqrt{8y} \right)}^{3}}}.......(5)$$ Now substituting x equal to 0 in equation (5) we get, $$\Rightarrow \dfrac{\sqrt{2y}}{{{\left( \sqrt{8y}+\sqrt{8y} \right)}^{3}}}.......(6)$$ Now adding similar terms in denominator and cubing it in equation (6) we get, $$\begin{aligned} & \Rightarrow \dfrac{\sqrt{2y}}{{{\left( 2\sqrt{8y} \right)}^{3}}} \\\ & \Rightarrow \dfrac{\sqrt{2}\sqrt{y}}{8\times 8\sqrt{8}\times y\sqrt{y}}..........(7) \\\ \end{aligned}$$ Canceling similar terms in equation (7) we get our answer, $$\Rightarrow \dfrac{1}{8\times 8\times 2\times y}=\dfrac{1}{128y}$$ Hence the answer is $$\dfrac{1}{128y}$$. So the correct answer is option (d). Note: Remembering the formula of square is important. We in a hurry can make a mistake in taking out common terms from equation (3) so we need to be careful while doing this. We also need to be vigilant while cubing the denominator in equation (6).