The function f(x) = (ln(1+ax) - ln(1-bx))/x is not defined at x = 0. The value which should be assig

Q: Wait, why do we reject the negative root here? Or is there no root?

Great question! In this specific problem, we are not solving a quadratic equation, so there are no roots to reject. We are evaluating a limit to find the value that 'plugs the hole' in the function. The result $a+b$ is the unique value that ensures the function is continuous at $x=0$.

Q: Can I just use L'Hôpital's Rule instead of the standard limit?

Absolutely! L'Hôpital's Rule is a powerful tool. If you differentiate the numerator and denominator with respect to $x$, you get $\frac{\frac{a}{1+ax} - \frac{-b}{1-bx}}{1}$. Evaluating this at $x=0$ gives $\frac{a}{1} + \frac{b}{1} = a+b$. Both methods are perfectly valid, but mastering the standard limit is faster and builds better intuition for competitive exams.

Q: Why is the second term $-b$ and not $+b$? I keep getting confused with the signs.

That is a classic trap! The original expression is $\ln(1-bx)$. To use the standard limit $\lim_{t \to 0} \frac{\ln(1+t)}{t} = 1$, we need the denominator to be $-bx$. So, we multiply by $\frac{-b}{-b}$. This leaves a $-b$ factor outside, which is why the second term becomes $-(-b) = +b$. Always keep the sign attached to the variable inside the log!

Question

The function $f(x) = \frac{\ln(1+ax) - \ln(1-bx)}{x}$ is not defined at $x = 0$. The value which should be assigned to $f$ at $x = 0$ so that it is continuous at $x = 0$, is

Options:

(A) $a - b$

(B) $a + b$

(C) $\ln a - \ln b$

(D) none of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are walking along the path of a function, $f(x) = \frac{\ln(1+ax) - \ln(1-bx)}{x}$. Everything seems smooth until you reach $x = 0$.

Suddenly, the path vanishes. You try to step on it, but you find yourself staring into an abyss—a classic indeterminate form, $\frac{0}{0}$.

In the world of calculus, this is not a dead end; it is a challenge. Our mission is to find the exact height of this 'hole' so we can fill it and make the function continuous.

### The Condition for Continuity

To make a function continuous at a point, the value of the function at that point must be perfectly aligned with the limit as we approach it. Mathematically, we require $f(0) = \lim_{x 	o 0} f(x)$.

Our objective is clear: evaluate the limit:

$$L = \lim_{x 	o 0} \frac{\ln(1+ax) - \ln(1-bx)}{x}$$

### The Toolkit: The Standard Logarithmic Limit

Every JEE aspirant needs a sharp toolkit. For logarithmic limits, our most powerful weapon is the standard form:

$$\lim_{t 	o 0} \frac{\ln(1+t)}{t} = 1$$

This formula is elegant because it tells us that as long as the argument inside the natural log matches the denominator, the limit will always collapse to unity. Our task is to force our expression to look like this.

### The Algebraic Dance

Let us break the fraction apart. Using the linearity of limits, we can write:

$$L = \lim_{x 	o 0} \frac{\ln(1+ax)}{x} - \lim_{x 	o 0} \frac{\ln(1-bx)}{x}$$

Now, look at the first term: $\frac{\ln(1+ax)}{x}$. The argument is $ax$, but the denominator is just $x$. We need an $a$ in the denominator.

So, we multiply and divide by $a$:

$$a \cdot \lim_{x 	o 0} \frac{\ln(1+ax)}{ax}$$

As $x$ approaches $0$, $ax$ also approaches $0$. This perfectly matches our standard limit, giving us $a \cdot (1) = a$.

Now, for the second term: $\frac{\ln(1-bx)}{x}$. The argument is $-bx$. We need $-bx$ in the denominator.

We multiply and divide by $-b$:

$$-b \cdot \lim_{x 	o 0} \frac{\ln(1-bx)}{-bx}$$

Again, as $x$ approaches $0$, $-bx$ approaches $0$. This gives us $-b \cdot (1) = -b$.

### The Grand Finale

Combining these results, we have:

$$L = a - (-b) = a + b$$

We have successfully navigated the indeterminate form. By defining **$f(0) = a + b$**, we effectively plug the hole in our graph, creating a seamless, continuous path.

This is the beauty of calculus—turning a point of confusion into a point of clarity. Keep practicing, stay curious, and remember that every hole in a function is just an opportunity to define something beautiful.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: The function $f (x) = \frac{l n ( 1 + a x ) - l n ( 1 - b x )}{x}$ is not defined at $x = 0$ . The value which should be assigned to $f$ at $x = 0$ so that it is continuous at $x = 0$ , is

Select Answer:

Visualized Solution

Visualizing the Discontinuity at $x = 0$

The Condition for Continuity

The Standard Logarithmic Limit

Splitting the Fraction

Adjusting the First Term

Adjusting the Second Term

Evaluating Both Limits

Filling the Hole for Continuity

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Condition for Continuity

The Toolkit

The Algebraic Dance

The Grand Finale

Similar Questions

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Select Answer:

Visualized Solution

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

The Condition for Continuity

The Standard Logarithmic Limit

Splitting the Fraction

Adjusting the First Term

Adjusting the Second Term

Evaluating Both Limits

Filling the Hole for Continuity

The Sigma Insight: Continuity at a Point and in an Interval

Analyzing the Setup

The Condition for Continuity

The Toolkit

The Algebraic Dance

The Grand Finale

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