Question

lim x tends to a
root 2x - root 3x-a divide by
root x - root a

Answer 1

Step-by-step explanation:

Given:$lim_{\: x \to a} \: \frac{ \sqrt{2x} - \sqrt{3x - a} }{ \sqrt{x} - \sqrt{a} }$

To find: Evaulate the limit.

Solution:

L'Hospital's Rule: If a rational function converts into 0/0 or ∞/∞ form after applying limit then Differentiate numerator and denominator separately(* Don't apply U/V form) and apply limit.

Step 1: Checking for the form.

First check whether the form is 0/0 or ∞/∞.

Put x=a in the function

$= \frac{ \sqrt{2a} - \sqrt{3a - a} }{ \sqrt{a} - \sqrt{a} }$

or

$= \frac{ \sqrt{2a} - \sqrt{2a} }{ \sqrt{a} - \sqrt{a} }$

$= \frac{0}{0}$

Step 2: Do differentiation.

As the function is

$lim_{x \to a} \: \frac{ \sqrt{2x} - \sqrt{3x - a} }{ \sqrt{x} - \sqrt{a} }$

Do differentiation

$lim_{x \to a} \: \left(\frac{ \frac{1}{ \sqrt{2} \sqrt{x} } - \frac{3}{2 \sqrt{3x - a} } }{ \frac{1}{2 \sqrt{x} } - 0 } \right)$

Apply limit

$= \: \left(\frac{ \frac{1}{ \sqrt{2} \sqrt{a} } - \frac{3}{2 \sqrt{3a - a} } }{ \frac{1}{2 \sqrt{a} } } \right)$

or

$= \: \left(\frac{ \frac{1}{ \sqrt{2} \sqrt{a} } - \frac{3}{2 \sqrt{2} \sqrt{a} } }{ \frac{1}{2 \sqrt{a} } } \right)$

take 1/√a common from both numerator and denominator and cancel

$= \: \left(\frac{ \frac{1}{ \sqrt{2} } - \frac{3}{2 \sqrt{2} } }{ \frac{1}{2} } \right)$

simplify

$= \left( \frac{\frac{2 - 3}{2 \sqrt{2}}}{ \frac{1}{2} } \right)$

or

$= \frac{ - 1}{2 \sqrt{2} } \times \frac{2}{1}$

or

$= \frac{ - 1}{ \sqrt{2} }$

Final answer:

$\bf lim_{\: x \to a} \: \frac{ \sqrt{2x} - \sqrt{3x - a} }{ \sqrt{x} - \sqrt{a} } = \frac{ - 1}{ \sqrt{2} }$

Hope it will help you.

Note*: Alternatively it can be solved by Taylor series expansion.

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