Evaluate lim (x→a) (√(a + 2x) - √(3x)) / (√(3a + x) - 2√x), (a ≠ 0).

Q: Can I just use L'Hopital's Rule here instead of all this rationalization?

That is a fantastic question! Yes, L'Hopital's Rule is a powerful tool in your arsenal. However, when dealing with square roots, the chain rule can lead to messy derivatives that are prone to calculation errors. Rationalization is often more elegant, algebraically cleaner, and helps you develop a deeper intuition for how these functions behave near the limit point. In JEE Advanced, mastering both methods is key, but rationalization is the 'surgical' approach here.

Q: Why can we cancel the term $(a - x)$? Isn't it zero?

This is a classic trap! Remember the fundamental definition of a limit: when we say $x \to a$, we mean that $x$ gets arbitrarily close to $a$, but $x$ is never *exactly* equal to $a$. Because $x \neq a$, the term $(a - x)$ is never zero. Since it is a non-zero value, we are perfectly justified in canceling it from both the numerator and the denominator.

Q: How do I know when to use double rationalization?

Great intuition! You should look for the 'double rationalization' strategy whenever you encounter a limit that results in the $0/0$ indeterminate form and contains square roots (or other radicals) in both the numerator and the denominator. If you see radicals in only one part, single rationalization is enough. If they are in both, you must neutralize both to uncover the common factor.

Question

Evaluate $\lim_{x 	o a} \frac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}}, (a 
eq 0).$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with evaluating the limit:
$$\lim_{x 	o a} \frac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}}$$

When dealing with limits, the first step is always to test the waters via direct substitution. Replacing $x$ with $a$ in the numerator yields $\sqrt{a + 2a} - \sqrt{3a} = \sqrt{3a} - \sqrt{3a} = 0$.

Similarly, the denominator becomes $\sqrt{3a + a} - 2\sqrt{a} = \sqrt{4a} - 2\sqrt{a} = 2\sqrt{a} - 2\sqrt{a} = 0$. We have arrived at the **$0/0$ indeterminate form**, which indicates a hidden factor causing the zero.

### The Strategy: Double Rationalization

Since we have radicals in both the numerator and the denominator, we must perform a double rationalization. We multiply the expression by the conjugates of both the numerator and the denominator:
$$\lim_{x 	o a} \left( \frac{\sqrt{a + 2x} - \sqrt{3x}}{\sqrt{3a + x} - 2\sqrt{x}} ight) \cdot \left( \frac{\sqrt{a + 2x} + \sqrt{3x}}{\sqrt{a + 2x} + \sqrt{3x}} ight) \cdot \left( \frac{\sqrt{3a + x} + 2\sqrt{x}}{\sqrt{3a + x} + 2\sqrt{x}} ight)$$

Applying the difference of squares identity, $(A - B)(A + B) = A^2 - B^2$, we simplify the numerator:
$$(\sqrt{a + 2x})^2 - (\sqrt{3x})^2 = a + 2x - 3x = a - x$$

Next, we simplify the denominator:
$$(\sqrt{3a + x})^2 - (2\sqrt{x})^2 = 3a + x - 4x = 3a - 3x = 3(a - x)$$

### Final Calculation

The expression now contains the common factor $(a - x)$ in both the numerator and the denominator. Since $x 	o a$, we know $x 
eq a$, allowing us to cancel the term:
$$\lim_{x 	o a} \frac{\sqrt{3a + x} + 2\sqrt{x}}{3(\sqrt{a + 2x} + \sqrt{3x})}$$

Now, we perform direct substitution by setting $x = a$:
$$	ext{Numerator: } \sqrt{3a + a} + 2\sqrt{a} = \sqrt{4a} + 2\sqrt{a} = 2\sqrt{a} + 2\sqrt{a} = 4\sqrt{a}$$
$$	ext{Denominator: } 3(\sqrt{a + 2a} + \sqrt{3a}) = 3(\sqrt{3a} + \sqrt{3a}) = 3(2\sqrt{3a}) = 6\sqrt{3}\sqrt{a}$$

The $\sqrt{a}$ terms cancel out, leaving us with:
$$\frac{4}{6\sqrt{3}} = \frac{2}{3\sqrt{3}}$$

The final result is **$\frac{2}{3\sqrt{3}}$**.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Evaluate $lim_{x \to a} \frac{a + 2 x - 3 x}{3 a + x - 2 x}, (a \neq = 0) .$

Visualized Solution

Introduction to the Limit

Checking the Form

Evaluating the Numerator

Evaluating the Denominator

The $\frac{0}{0}$ Indeterminate Form

Double Rationalization Setup

Applying the Difference of Squares

Simplifying the Numerator

Simplifying the Denominator

Combining Like Terms

Canceling the Common Factor

Final Substitution

Evaluating the Result

Final Simplification

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

The Strategy

Final Calculation

Similar Questions

$lim_{x \to a} \frac{( a + 2 x ) ^{1/3} - ( 3 x ) ^{1/3}}{( 3 a + x ) ^{1/3} - ( 4 x ) ^{1/3}}$ $(a \neq = 0)$ is equal to

If the value of $lim_{x \to 0} (2 - cos x cos 2 x)^{\frac{x + 2}{x ^{2}}}$ is equal to $e^{a}$ , then $a$ is equal to

$lim_{x \to \infty} \frac{( 2 x ^{2} - 3 x + 5 ) ( 3 x - 1 ) ^{\frac{x}{2}}}{( 3 x ^{2} + 5 x + 4 ) ( 3 x + 2 ) ^{x}}$ is equal to:

$lim_{x \to 0} (\frac{( x + 2 c o s x ) ^{3} + 2 ( x + 2 c o s x ) ^{2} + 3 s i n ( x + 2 c o s x )}{( x + 2 ) ^{3} + 2 ( x + 2 ) ^{2} + 3 s i n ( x + 2 )})^{\frac{100}{x}}$ is equal to ______.

$lim_{x \to 2} \frac{3 ^{x} + 3 ^{3 - x} - 12}{3 ^{- x /2} - 3 ^{1 - x}}$ is equal to.....

For $x \in R, lim_{x \to \infty} (\frac{x - 3}{x + 2})^{x} =$

The value of the limit $lim_{x \to \frac{π}{2}} \frac{4 2 ( s i n 3 x + s i n x )}{( 2 s i n 2 x s i n \frac{3 x}{2} + c o s \frac{5 x}{2} ) - ( 2 + 2 c o s 2 x + c o s \frac{3 x}{2} )}$ is ____.

$lim_{x \to 0} \frac{e - ( 1 + 2 x ) ^{\frac{1}{2 x}}}{x}$ is equal to

Find $lim_{x \to 0} {tan (π /4 + x)}^{1/ x} .$

$lim_{x \to 2} \frac{3 ^{x} + 3 ^{3 - x} - 12}{3 ^{- x /2} - 3 ^{1 - x}}$ is equal to

Animated Solution for Mathematics - Limits, Continuity and Differentiability: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate limx→a​3a+x​−2x​a+2x​−3x​​,(a=0).