Question

If \(x_{1}\) and \(x_{2}\) are the real and distinct roots of \(a x^{2}+b x+c=0\), then \(\operatorname{Lim}_{x \rightarrow x_{1}}\left(1+\sin \left(a x^{2}+b x+c\right)\right)^{\frac{1}{x-x_{1}}}\) is equal to (1) \(e^{x_{1}-x_{2}}\) (2) \(e^{x_{2}-x_{1}}\) (3) \(e^{a\left(x_{1}-x_{2}\right)}\) (4) \(e^{a\left(x_{2}-x_{1}\right)}\)

Short answer

1

Step-by-step solution

Identify the roots

Given the quadratic equation is of the form \(ax^{2} + bx + c = 0\) the real and distinct roots are denoted by \(x_{1}\) and \(x_{2}\)

Approach the limit

We need to evaluate the limit: \(\text{Lim}_{x \rightarrow x_{1}}(1+\text{sin} (a x^{2}+b x+c))^{\frac{1}{x-x_{1}}}\). Notice the term \(ax^{2} + bx + c = 0\) when \(x = x_ {1}\).

Substitute root

Since \(x_{1}\) is a root, then \(a x_{1}^{2} + b x_{1} + c = 0\). Thus, \(a x^{2} + b x + c \rightarrow 0\) as \(x \rightarrow x_{1}\). The limit simplifies to \(\text{Lim}_{x \rightarrow x_{1}}(1+\text{sin} (0))^{\frac{1}{x-x_{1}}}\).

Simplify the Limit expression

We know that \(\text{sin}(0) = 0\), so the expression becomes \(\text{Lim}_{x \rightarrow x_{1}}(1+0)^{\frac{1}{x-x_{1}}}\) which is \(\text{Lim}_{x \rightarrow x_{1}}1^{\frac{1}{x-x_{1}}}\).

Evaluate the final limit

Since \(1\) raised to any power is still \(1\), the expression evaluates to \(1\). Hence, \(\text{Lim}_{x \rightarrow x_{1}}(1+\text{sin}(a x^{2}+b x+c))^{\frac{1}{x-x_{1}}} = 1\).

Key concepts

Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\) where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). These equations can have different types of roots depending on the discriminant \( \triangle = b^2 - 4ac \). The roots can be:

• Real and distinct if \( \triangle > 0 \)
• Real and equal if \( \triangle = 0 \)
• Complex if \( \triangle < 0 \)

For this exercise, we are working with real and distinct roots, meaning our discriminant is positive. You can find the roots using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] These roots are denoted by \(x_1\) and \(x_2\), and they are the values that satisfy the equation \( ax^2 + bx + c = 0 \).

Limit Evaluation

In calculus, evaluating limits helps us determine the value that a function approaches as the input approaches some value. The limit in the exercise is \( \text{Lim}_{x \rightarrow x_{1}}(1+\text{sin}(ax^2+bx+c))^{\frac{1}{x-x_1}} \). To evaluate this limit:

• Notice that \( ax^2 + bx + c = 0\) when \( x = x_1 \) because \( x_1 \) is a root.
• Substituting \( x_1 \) into the quadratic equation simplifies the term \(ax^2+bx+c \) to zero.
• The limit then converts to \( \text{Lim}_{x \rightarrow x_{1}}(1+\text{sin}(0))^{\frac{1}{x-x_1}} \).
• Given \( \text{sin}(0) = 0 \), this simplifies further to \( \text{Lim}_{x \rightarrow x_{1}}1^{\frac{1}{x-x_1}} \).
• The expression evaluates to \(1\) since any number raised to the power of zero is \(1\).

This step-by-step simplification helps you grasp the concept of limit evaluation in this context.

Real and Distinct Roots

Real and distinct roots are the solutions to the quadratic equation that are both real numbers and different from each other. These roots are particularly important in problems involving quadratic equations where you need to explore their properties more deeply. For example:

• In solving \(ax^2 + bx + c = 0\), if the discriminant \( \triangle = b^2 - 4ac \) is greater than zero, it indicates that the quadratic equation has two real and distinct roots.
• These roots can be computed using the quadratic formula and are symbolized as \(x_1 \) and \(x_2 \).
• When these roots are substituted into other mathematical expressions or limits, especially in calculus problems like the one in the exercise, they help simplify and solve the expressions effectively.

Understanding the characteristics of real and distinct roots not only aids in solving quadratic equations but also simplifies complex calculus problems.