Problem 18 - Single Variable Calculus

Question

Find the limit. $$ \lim _{x \rightarrow(\pi / 2)^{-}} \frac{2 e^{\operatorname{lan} x}}{2 x-\pi} $$

Step-by-Step Solution

Answer

The limit of the given expression as x approaches \(\pi/2\) from the left side is 1.

Step 1: Simplify the expression

First, we need to simplify the expression. The expression inside the limit is given by: \[ \frac{2 e^{\operatorname{lan} x}}{2 x-\pi} \] e^(lan x) is same as x, hence the expression becomes: \[ \frac{2x}{2x-\pi} \] Now, let's find the limit of this simplified expression as x approaches \(\pi/2\) from the left side.

Step 2: Apply L'Hôpital's Rule

Since both numerator and denominator go to zero when x approaches \(\pi/2\), it's a 0/0 indeterminate form. Therefore, we can apply L'Hôpital's Rule which states that: \[ \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)} \] if the limit on the right exists and provided that g'(x) ≠ 0 in a deleted neighborhood of c. In our case, f(x) = 2x and g(x) = 2x - \(\pi\). Now let's find f'(x) and g'(x). f'(x) = \(\frac{d}{dx}(2x) = 2\) g'(x) = \(\frac{d}{dx}(2x-\pi) = 2\) Now let's find the limit of their quotient as x approaches \(\pi/2\) from the left side: \[ \lim _{x \rightarrow(\pi / 2)^{-}} \frac{2}{2} \]

Step 3: Calculate the limit

Finally, as both the numerator and the denominator are constant, the limit is: \[\lim _{x \rightarrow(\pi / 2)^{-}} \frac{2}{2} = \frac{2}{2} = 1\] So, the limit of the given expression as x approaches \(\pi/2\) from the left side is 1.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a very useful tool in calculus for finding limits that would otherwise be challenging to evaluate directly. When you encounter a limit where both the numerator and the denominator tend to zero as a variable approaches a certain value, this results in an indeterminate form. L'Hôpital's Rule provides a systematic approach to evaluate this type of limit.

To apply L'Hôpital's Rule, you first need to check if the limit of the original function results in a \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) indeterminate form. If so, you can differentiate both the numerator and the denominator separately. Then, you evaluate the limit of the new function created by the derivatives. This process can sometimes be repeated if the resulting expression still presents an indeterminate form.

Remember, the rule only applies when certain conditions are met: both functions must be differentiable around the point of interest and the derivative of the denominator should not be zero around that point.

Indeterminate Forms

In calculus, indeterminate forms appear frequently, especially when evaluating limits. Common indeterminate forms include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \(0 \times \infty\), \(\infty - \infty\), and others.

These forms signal that straightforward substitution will not yield a definite answer. Hence, special techniques like L'Hôpital's Rule, factoring, or algebraic manipulation are necessary to tackle these problems.

• The expression \( \frac{0}{0} \) typically points to a subtle interplay between the numerator and the denominator \ that might simplify upon differentiating.
• Similarly, \( \frac{\infty}{\infty} \) suggests that both parts are growing unbounded \ and may cancel each other under differentiation.

Mastering the understanding and handling of indeterminate forms is crucial, as it \ allows the resolution of seemingly complicated problems by simplifying them.

Differentiation

Differentiation is a fundamental concept in calculus involving the computation of a derivative. The derivative measures how a function changes as its input changes.

For instance, if you have a function \( y = f(x) \), the derivative, denoted \( f'(x) \) or \( \frac{d}{dx}f(x) \), \ defines the rate at which \( y \) changes with respect to \( x \).

In the context of L'Hôpital's Rule, differentiating both the numerator and denominator of a function can resolve an indeterminate form. When you differentiate a linear function like \( 2x \), the result is a constant \( 2 \). This is because linear functions have a constant rate of change.

• Differentiation relies on the rules like the power rule \( \frac{d}{dx}[x^n] = nx^{n-1} \).
• For trigonometric, exponential, or other special functions, \ derivatives may involve more specific rules or formulations.

Understanding how to differentiate accurately is essential in applying L'Hôpital's Rule effectively.

Evaluating Limits

Evaluating limits in calculus involves determining the value that a function approaches as its input approaches a certain value. This concept is fundamental because it sets the foundation for defining derivatives and integrals.

To find a limit, one common approach is to directly substitute the value into the function. However, in cases of indeterminate forms, as discussed earlier, further techniques such as L'Hôpital's Rule come into play.

For example, if evaluating \[\lim _{x \rightarrow(\pi / 2)^{-}} \frac{2 e^{\operatorname{lan} x}}{2x-\pi}\]This initially seems complicated until we simplify it and discover the indeterminate form.

• The limit was simplified to \( \lim _{x \rightarrow(\pi / 2)^{-}} \frac{2x}{2x-\pi} \), which forms \( \frac{0}{0} \), allowing the use of L'Hôpital's Rule.
• After differentiating, \ the limit turned into a constant expression \( \lim _{x \rightarrow(\pi / 2)^{-}} \frac{2}{2} = 1 \), which is easy to evaluate.

Learning to evaluate limits effectively encompasses knowing when and how to apply different calculus methods to make the problem straightforward.