If α, β are the distinct roots of x² + bx + c = 0, then lim(x→β) [e^(2(x² + bx + c)) - 1 - 2(x² + bx

Q: Why can't we just use L'Hopital's rule?

While L'Hopital's rule is a powerful tool, applying it twice to an expression involving an exponential function and a quadratic can lead to a messy chain rule nightmare. The Taylor series approach is much cleaner and reveals the underlying structure of the function.

Q: How do I know when to use the Taylor series expansion?

Whenever you see an expression of the form $\frac{e^u - 1 - u}{u^2}$ where $u \to 0$, the Taylor series is your best friend. It allows you to replace the transcendental function with a simple polynomial, making the limit evaluation straightforward.

Q: Is it always valid to replace $f(x)$ with $(x-\alpha)(x-\beta)$?

Yes, absolutely! The Factor Theorem is a fundamental property of polynomials. Since $\alpha$ and $\beta$ are roots of $x^2 + bx + c = 0$, the expression must be divisible by $(x-\alpha)$ and $(x-\beta)$. This is a powerful way to simplify expressions involving roots.

Question

If $\alpha, \beta$ are the distinct roots of $x^2 + bx + c = 0$, then $\lim_{x 	o \beta} \frac{e^{2(x^2 + bx + c)} - 1 - 2(x^2 + bx + c)}{(x - \beta)^2}$ is equal to:

Options:

(A) $b^2 + 4c$

(B) $2(b^2 + 4c)$

(C) $2(b^2 - 4c)$

(D) $b^2 - 4c$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before the graph of a quadratic function, $f(x) = x^2 + bx + c$. It is a classic parabola, sweeping gracefully across the Cartesian plane.

The problem states that $\alpha$ and $\beta$ are its distinct roots. Geometrically, this means the parabola intersects the $x$-axis at exactly two points: $\alpha$ and $\beta$. At these precise locations, the function's value is zero.

By the Factor Theorem, we can express this quadratic not just as a sum of terms, but as a product of its roots:
$$f(x) = (x - \alpha)(x - \beta)$$

This form is powerful because it allows us to see exactly what happens to the function as $x$ approaches one of its roots.

### The Limit's Anatomy: Unveiling the Taylor Series

Now, let us turn our attention to the limit:
$$\lim_{x 	o \beta} \frac{e^{2f(x)} - 1 - 2f(x)}{(x - \beta)^2}$$

As $x$ slides along the $x$-axis toward $\beta$, the value of $f(x)$ drops toward zero. This means the exponent $2f(x)$ is also approaching zero.

This is a classic setup for a Taylor series expansion. Whenever you see $e^u - 1 - u$ where $u 	o 0$, you should recall the Maclaurin series for $e^u$:
$$e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$$

By setting $t = 2f(x)$, we can rewrite our numerator. As $x 	o \beta$, $t$ also approaches zero. The expansion becomes $e^t \approx 1 + t + \frac{t^2}{2}$.

### The Elegant Cancellation

Let us substitute this back into our numerator:
$$e^{2f(x)} - 1 - 2f(x) \approx \left(1 + 2f(x) + \frac{(2f(x))^2}{2}ight) - 1 - 2f(x)$$

Notice the magic: the $1$ and the $2f(x)$ terms cancel out perfectly with the $-1$ and $-2f(x)$ already present. We are left with $\frac{(2f(x))^2}{2}$, which simplifies to $2(f(x))^2$.

Our limit has now reduced to a much friendlier form:
$$\lim_{x 	o \beta} \frac{2(f(x))^2}{(x - \beta)^2}$$

### The Bridge to Coefficients

Now, we bring back our factorized form: $f(x) = (x - \alpha)(x - \beta)$. Substituting this into our limit, we get:
$$\lim_{x 	o \beta} \frac{2(x - \alpha)^2(x - \beta)^2}{(x - \beta)^2}$$

Since $x$ is approaching $\beta$ but is not equal to $\beta$, we can safely cancel the $(x - \beta)^2$ terms. We are left with $\lim_{x 	o \beta} 2(x - \alpha)^2$, which evaluates simply to $2(\beta - \alpha)^2$.

Finally, we connect this back to the coefficients $b$ and $c$. Using Vieta's formulas, we know $\alpha + \beta = -b$ and $\alpha\beta = c$.

The identity $(\beta - \alpha)^2 = (\alpha + \beta)^2 - 4\alpha\beta$ allows us to write this as $b^2 - 4c$. Multiplying by the $2$ we had outside, our **final answer** is:
$$2(b^2 - 4c)$$

Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $α, β$ are the distinct roots of $x^{2} + b x + c = 0$ , then $lim_{x \to β} \frac{e ^{2 (x^{2} + b x + c)} - 1 - 2 ( x ^{2} + b x + c )}{( x - β ) ^{2}}$ is equal to:

Select Answer:

Visualized Solution

Define $f (x) = x^{2} + b x + c$

Factorize $f (x)$

Analyze the Limit Structure

Apply Taylor Expansion

Simplify the Numerator

Substitute $f (x)$ into Limit

Cancel the Common Factor

Evaluate the Limit

Relate Roots to Coefficients

Final Calculation

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

Analyzing the Setup

The Limit's Anatomy

The Elegant Cancellation

The Bridge to Coefficients

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