Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0^{-}} \frac{\sin x+\tan x}{e^{x}+e^{-x}-2} $$

Short answer

The limit is \(-\infty\).

Step-by-step solution

Evaluate the Limit Directly

First, we substitute the value \( x = 0^- \) into the expression \( \frac{\sin x + \tan x}{e^{x} + e^{-x} - 2} \). Calculating each part: \( \sin(0) = 0 \), \( \tan(0) = 0 \), \( e^{0} = 1 \), \( e^{-0} = 1 \). Thus, the expression evaluates to \( \frac{0 + 0}{1 + 1 - 2} = \frac{0}{0} \), which is indeterminate.

Apply L'Hôpital's Rule

Since we have an indeterminate form \( \frac{0}{0} \), we can apply L'Hôpital's Rule. This involves differentiating the numerator and denominator separately. Differentiating, we get \( \cos x + \sec^2 x \) for the numerator and \( e^x - e^{-x} \) for the denominator.

Differentiate and Re-Evaluate the Limit

We substitute the differentiated expressions into the limit: \( \lim_{x \rightarrow 0^-} \frac{\cos x + \sec^2 x}{e^x - e^{-x}} \). Substituting \( x = 0^- \): \( \cos 0 = 1 \), \( \sec^2 0 = 1 \), \( e^0 = 1 \), and \( e^{-0} = 1 \), which gives \( \frac{1 + 1}{1 - 1} = \frac{2}{0} \). Since \( x \rightarrow 0^- \), \( e^x - e^{-x} \rightarrow 0^- \), this results in an expression that behaves like a division by a very small negative number leading to \(-\infty\).

Key concepts

Understanding Indeterminate Forms

When solving limit problems in calculus, you might encounter expressions that seem tricky. Some of these are called "indeterminate forms." An indeterminate form happens when the usual operations of arithmetic (like addition, subtraction, multiplication, and division) don't work as expected. For example, in the limit \[\lim_{x \to a} \frac{f(x)}{g(x)}\]if substituting \(x = a\) directly results in \(\frac{0}{0}\), this is an indeterminate form. It's not defined because dividing zero by zero doesn't give a clear answer. Unlike a regular division by zero, where the result might be undefined or infinite, these forms require special methods to resolve.Here are some common indeterminate forms:

• \(\frac{0}{0}\)
• \(\frac{\infty}{\infty}\)
• \(0 \times \infty\)
• \(\infty - \infty\)

Identifying these forms is crucial because they indicate that normal algebraic methods aren't enough. Instead, we use techniques like L'Hôpital's Rule to find limits.

The Process of Limit Evaluation

Evaluating limits is like solving a puzzle, where you find the value a function approaches as the input approaches a certain point. It's a fundamental concept in calculus that helps us understand behaviors of functions near specific values.To evaluate the limit, follow these steps:

• Direct Substitution: Start by plugging the point into the function. If it gives a clear value, that's your answer!
• Check for Indeterminate Forms: If direct substitution results in forms like \(\frac{0}{0}\), you have to rethink your approach.
• Use L'Hôpital's Rule: When faced with \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), differentiate the numerator and the denominator separately and then evaluate the limit again.

Limit evaluation helps in understanding how functions behave, especially when dealing with potential discontinuities or where a function isn't defined at a certain point.

Solving Calculus Problems with L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus problem solving. When faced with an indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), this rule provides a way to simplify and evaluate limits easily. Here's how to apply L'Hôpital's Rule:

• Differentiate the numerator and denominator separately.
• Evaluate the limit of the new function.

It's important to note that L'Hôpital's Rule only applies when the original limit yields an indeterminate form. In the example provided, differentiating \(\sin x + \tan x\) gives \(\cos x + \sec^2 x\), and differentiating \(e^x + e^{-x} - 2\) results in \(e^x - e^{-x}\).Calculus problem solving often includes several trails of limit evaluation, ensuring indeterminate forms are addressed correctly. Mastering these techniques can make resolving complex calculus questions less daunting.

Differentiation for Simplifying Limits

Differentiation is a key calculus operation that calculates the rate of change of a function. When evaluating limits involving indeterminate forms, differentiation allows us to transform the original problem into something simpler. To differentiate a function, you apply specific rules like the power rule, product rule, quotient rule, or chain rule. In the context of L'Hôpital's Rule, you apply differentiation directly to the numerator and denominator.For example, converting \(\sin x + \tan x\) into its derivative involves using:

• \(\cos x\) for \(\sin x\)
• \(\sec^2 x\) for \(\tan x\)

Similarly, differentiating \(e^x + e^{-x} - 2\) gives \(e^x - e^{-x}\).Differentiation helps turn an originally difficult limit problem into a straightforward calculation, making it easy to find the value that the function approaches.