Let the function f(x) = { (log_e(1 + 5x) - log_e(1 + αx)) / x, if x ≠ 0; 10, if x = 0 } be continuou

Q: Why can't we just plug in $x=0$ directly into the expression?

If you try to plug in $x=0$, you get $\frac{\ln(1) - \ln(1)}{0}$, which is $\frac{0}{0}$. This is an indeterminate form! It doesn't mean the limit doesn't exist; it means we need to use calculus tools like standard limits to find the 'hidden' value.

Q: Why does the limit $\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$ work?

Think of the Taylor series expansion for $\ln(1+u) \approx u - \frac{u^2}{2} + \dots$. When you divide by $u$, you get $1 - \frac{u}{2} + \dots$. As $u$ approaches zero, all terms except the $1$ vanish. That's the beauty of it!

Q: Is it possible for $\alpha$ to be a negative number?

Absolutely! In this problem, we found $\alpha = -5$. Continuity doesn't care about the sign; it only cares that the limit matches the function value. Always trust the algebra!

Question

Let the function $f(x) = \begin{cases} \frac{\log_e(1 + 5x) - \log_e(1 + \alpha x)}{x}, & 	ext{if } x 
eq 0 \ 10, & 	ext{if } x = 0 \end{cases}$ be continuous at $x = 0$. The $\alpha$ is equal to :

Options:

(A) 10

(B) -10

(C) 5

(D) -5

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the JEE journey! Today, we explore a problem that serves as a beautiful demonstration of the concept of continuity.

Our problem provides a piecewise function $f(x)$ defined as:

$$f(x) = \begin{cases} \frac{\ln(1 + 5x) - \ln(1 + \alpha x)}{x}, & x 
eq 0 \ 10, & x = 0 \end{cases}$$

Our mission is to find the value of $\alpha$ that makes this function perfectly continuous at $x=0$.

### The Condition for Continuity

Mathematically, a function is continuous at a point if the limit of the function as $x$ approaches that point equals the actual value of the function at that point.

We set up our governing equation:

$$\lim_{x 	o 0} f(x) = f(0)$$

Given $f(0) = 10$, our goal is to solve the following limit equation:

$$\lim_{x 	o 0} \frac{\ln(1 + 5x) - \ln(1 + \alpha x)}{x} = 10$$

### The Art of Manipulation

If we attempt to evaluate the limit by plugging in $x=0$ directly, we encounter the indeterminate form $\frac{0}{0}$. To resolve this, we utilize the standard limit property:

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

We split our expression into two manageable parts:

$$\lim_{x 	o 0} \frac{\ln(1 + 5x)}{x} - \lim_{x 	o 0} \frac{\ln(1 + \alpha x)}{x} = 10$$

To match the standard form, we multiply and divide the first term by $5$ and the second term by $\alpha$:

$$5 \cdot \lim_{x 	o 0} \frac{\ln(1 + 5x)}{5x} - \alpha \cdot \lim_{x 	o 0} \frac{\ln(1 + \alpha x)}{\alpha x} = 10$$

### The Final Resolution

As $x 	o 0$, both $5x$ and $\alpha x$ approach zero. Applying the standard limit property, both fractional terms evaluate to $1$.

The expression simplifies to:

$$5(1) - \alpha(1) = 10$$

This yields the linear equation:

$$5 - \alpha = 10$$

Solving for $\alpha$, we find:

$$-\alpha = 5 \Rightarrow \mathbf{\alpha = -5}$$

By carefully manipulating the expression and applying the fundamental properties of limits, we have ensured the continuity of the function. Every complex problem is simply a collection of elegant steps waiting to be discovered.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let the function $f (x) = {\frac{l o g _{e} ( 1 + 5 x ) - l o g _{e} ( 1 + α x )}{x}, 10, if x \neq = 0 if x = 0$ be continuous at $x = 0$ . The $α$ is equal to :

Select Answer:

Visualized Solution

The Piecewise Function

Condition for Continuity

Setting up the Equation

Standard Logarithmic Limit

Splitting the Limit

Adjusting the First Term

Adjusting the Second Term

Evaluating the Limits

Solving for $α$

Final Result

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

Analyzing the Setup

The Condition for Continuity

The Art of Manipulation

The Final Resolution

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