lim x→a ((a+2x)^(1/3)-(3x)^(1/3))/((3a+x)^(1/3)-(4x)^(1/3)) (a ≠ 0) is equal to
(A) (2/3)^(1/3) (2/3

Q: Wait, why do we use L'Hospital's Rule here? Can't we just rationalize the cube roots?

You absolutely could rationalize the cube roots, but it would be a nightmare of algebraic manipulation! L'Hospital's Rule is a powerful shortcut for $\frac{0}{0}$ indeterminate forms. It transforms a difficult limit problem into a derivative problem, which is often much cleaner. In JEE, efficiency is key, and L'Hospital's Rule is a tool you must have in your arsenal.

Q: I keep messing up the Chain Rule. Why do we multiply by the derivative of the inside term?

Think of the Chain Rule as a way to account for the 'speed' of the inner function. When you have a composite function like $(a+2x)^{1/3}$, the outer function (the cube root) is changing, but the inner function $(a+2x)$ is also changing at its own rate. Multiplying by the derivative of the inner term (which is $2$ in this case) is the mathematical way of 'chaining' these rates of change together. It is essential for accuracy!

Q: Why does the variable $a$ cancel out at the end?

That is the beauty of these problems! The variable $a$ is just a parameter. When you evaluate the derivatives at $x=a$, the terms involving $a$ become proportional to each other. Because the limit is a ratio, these proportional factors cancel out, leaving you with a numerical constant. It is a sign that your derivation is on the right track.

Question

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

Options:

(A) $\left( \frac{2}{3} ight)^{1/3} \left( \frac{2}{3} ight)^{1/3}$

(B) $\left( \frac{2}{3} ight)^{1/3} \left( \frac{4}{3} ight)^{1/3}$

(C) $\left( \frac{2}{3} ight)^{1/3}$

(D) $\left( \frac{4}{3} ight)^{1/3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to tackle a problem that often intimidates students, but once you peel back the layers, it reveals a beautiful, logical structure. We are dealing with a limit involving cube roots.

In the world of JEE Advanced, the most complex-looking problems often yield to the most fundamental principles.

### The Indeterminate Trap

Imagine you are standing on the graph of the function $f(x) = (a+2x)^{1/3} - (3x)^{1/3}$. As $x$ approaches $a$, you are trying to find the value of the function.

If you simply plug in $x=a$, you get $(a+2a)^{1/3} - (3a)^{1/3}$, which is $(3a)^{1/3} - (3a)^{1/3} = 0$. The same happens in the denominator, resulting in the classic $\frac{0}{0}$ indeterminate form.

This is like a fog that hides the true value of the function. To clear this fog, we use **L'Hospital's Rule**. Geometrically, this rule tells us that when two functions intersect the $x$-axis at the same point, the limit of their ratio is simply the ratio of their slopes at that point.

### The Calculus Grind

Now, we need to find the slopes. This means we must differentiate. Let's focus on the numerator $f(x) = (a+2x)^{1/3} - (3x)^{1/3}$.

Applying the power rule and the Chain Rule, we get:

$$f'(x) = \frac{1}{3}(a+2x)^{-2/3} \cdot 2 - \frac{1}{3}(3x)^{-2/3} \cdot 3$$

We brought the power down, reduced it by one, and then multiplied by the derivative of the inner function. For $(a+2x)$, the derivative is $2$. For $(3x)$, the derivative is $3$.

Now, let's do the same for the denominator $g(x) = (3a+x)^{1/3} - (4x)^{1/3}$:

$$g'(x) = \frac{1}{3}(3a+x)^{-2/3} \cdot 1 - \frac{1}{3}(4x)^{-2/3} \cdot 4$$

### The Algebraic Cleanup

This is where the magic happens. We evaluate these derivatives at $x=a$:

$$f'(a) = \frac{1}{3}(3a)^{-2/3} \cdot 2 - \frac{1}{3}(3a)^{-2/3} \cdot 3 = \frac{1}{3}(3a)^{-2/3}(2-3) = -\frac{1}{3}(3a)^{-2/3}$$

$$g'(a) = \frac{1}{3}(4a)^{-2/3} \cdot 1 - \frac{1}{3}(4a)^{-2/3} \cdot 4 = \frac{1}{3}(4a)^{-2/3}(1-4) = -(4a)^{-2/3}$$

Notice how the terms simplify? The negative signs cancel out, and we are left with a ratio of powers. When we divide $f'(a)$ by $g'(a)$, the variable $a$ effectively disappears from the ratio.

### Final Calculation

The limit is calculated as follows:

$$	ext{Limit} = \frac{-\frac{1}{3}(3a)^{-2/3}}{-(4a)^{-2/3}} = \frac{1}{3} \cdot \frac{(4a)^{2/3}}{(3a)^{2/3}} = \frac{1}{3} \left( \frac{4}{3} ight)^{2/3}$$

We have navigated through the indeterminate form, applied the Chain Rule with precision, and simplified the expression. The final result is:

$$\mathbf{\frac{1}{3} \left( \frac{4}{3} ight)^{2/3}}$$

Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{x \to a} \frac{( a + 2 x ) ^{1/3} - ( 3 x ) ^{1/3}}{( 3 a + x ) ^{1/3} - ( 4 x ) ^{1/3}}$ $(a \neq = 0)$ is equal to

Select Answer:

Visualized Solution

Check the Indeterminate Form at $x = a$

Visualize the $\frac{0}{0}$ Form

Apply L'Hospital's Rule

Differentiate the Numerator $f (x)$

Execute Chain Rule for $f^{'} (x)$

Differentiate the Denominator $g (x)$

Execute Chain Rule for $g^{'} (x)$

Substitute $x = a$ in Derivatives

Simplify $f^{'} (a)$ and $g^{'} (a)$

Calculate Final Limit Ratio

Match with Options

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

Analyzing the Setup

The Indeterminate Trap

The Calculus Grind

The Algebraic Cleanup

Final Calculation

Similar Questions

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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; } limx→a​(3a+x)1/3−(4x)1/3(a+2x)1/3−(3x)1/3​ (a=0) is equal to

Select Answer:

Visualized Solution

Check the Indeterminate Form at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=a

Visualize the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 00​ Form

Apply L'Hospital's Rule

Differentiate the Numerator .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Execute Chain Rule for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

Differentiate the Denominator .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g(x)

Execute Chain Rule for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′(x)

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=a in Derivatives

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(a) and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } g′(a)

Calculate Final Limit Ratio

Match with Options

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

Analyzing the Setup

The Indeterminate Trap

The Calculus Grind

The Algebraic Cleanup

Final Calculation

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Animated Solution for Mathematics - Limits, Continuity and Differentiability: $lim_{x \to a} \frac{( a + 2 x ) ^{1/3} - ( 3 x ) ^{1/3}}{( 3 a + x ) ^{1/3} - ( 4 x ) ^{1/3}}$ $(a \neq = 0)$ is equal to

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