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Chapter 3: Problem 8

evaluate the limit using l'H么pital's Rule if appropriate. $$ \lim _{x \rightarrow 0} \frac{e^{x}-1}{x+\sin x} $$

Short Answer

Expert verified
The limit of the given function using l'H么pital's Rule is: \(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x+\sin x} = \frac{1}{2}\).

Step by step solution

01

Check if l'H么pital's Rule is applicable

We need to verify that both the numerator and the denominator approach 0 as x approaches 0: $$ \lim_{x \rightarrow 0} (e^x - 1) = e^0 - 1 = 1 - 1 = 0 $$ $$ \lim_{x \rightarrow 0} (x + \sin x) = 0 + \sin0 = 0 + 0 = 0 $$ Since both limits are 0, we can apply l'H么pital's Rule.
02

Find the derivatives of the numerator and the denominator

Next, we will find the first derivative of the numerator and the denominator with respect to x: The derivative of the numerator: $$ \frac{d}{dx}(e^x - 1) = e^x $$ The derivative of the denominator: $$ \frac{d}{dx}(x + \sin x) = 1 + \cos x $$
03

Apply l'H么pital's Rule

Now, we will apply l'H么pital's Rule by taking the limit of the ratio of the derivatives as x approaches 0: $$ \lim_{x \rightarrow 0} \frac{e^x}{1 + \cos x} $$
04

Evaluate the limit of the new function

We can evaluate the limit of the new function by plugging in x = 0: $$ \lim_{x \rightarrow 0} \frac{e^x}{1 + \cos x} = \frac{e^0}{1 + \cos 0} = \frac{1}{1 + 1} = \frac{1}{2} $$ The limit of the given function using l'H么pital's Rule is: $$ \lim _{x \rightarrow 0} \frac{e^{x}-1}{x+\sin x} = \frac{1}{2} $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Limits in Calculus
Limits are fundamental in calculus, providing the foundation for concepts like continuity and derivatives. They help us understand what a function approaches as the input gets closer to a specific value. In our exercise, we explored the limit as \(x\) approaches zero.

When applying limits, it's crucial to check if substitution gives a direct solution or leads to forms like \(\frac{0}{0}\). This allows us to decide if we need a different approach, such as l'H么pital's Rule. For example,
  • The limit \( \lim_{x \rightarrow 0} (e^x - 1) = 0 \)
  • The limit \( \lim_{x \rightarrow 0} (x + \sin x) = 0 \)
These equalities confirm a \(\frac{0}{0}\) indeterminate form, signaling the potential use of l'H么pital's Rule.
Basics of Derivatives
Derivatives measure how a function changes as its input changes, representing the rate of change or slope of a function at any point. They are key when using l'H么pital's Rule, where we differentiate the numerator and the denominator of an indeterminate form.

For the given exercise, we needed:
  • The derivative of \(e^x - 1\), which is \(e^x\) since the derivative of any constant is zero.
  • The derivative of \(x + \sin x\), which results in \(1 + \cos x\) as the derivative of \(x\) is 1 and that of \(\sin x\) is \(\cos x\).
With these derivatives, we simplify and find the limit of the new function as \(x\) approaches zero.
Understanding Exponential Functions
Exponential functions involve the constant \(e\), approximately 2.718, which arises naturally in various mathematical contexts, especially continuous growth processes. The term \(e^x\) showcases continuous compounding.

When examining the function \(e^x - 1\) for limits or derivatives:
  • The derivative \(e^x\) is unique as it remains the same even after differentiation.
  • They're central to calculus due to their predictable growth pattern and easy differentiation.
These properties make exponential functions a centerpiece in many calculus problems, streamlining processes like using l'H么pital's Rule.
Insights into Trigonometric Functions
Trigonometric functions, such as \(\sin x\) and \(\cos x\), derive from the geometry of triangles and are integral to various mathematical fields. They model periodic phenomena like waves or circular motion.

In the limit \( \lim_{x \rightarrow 0} \frac{e^x - 1}{x + \sin x} \), we used:
  • \(\sin 0 = 0\), emphasizing its value at the limit point.
  • \(\cos x\), appearing in derivatives, is pivotal for understanding changes in the sine function.
Trigonometric derivatives aid in modifying complex functions and resolving indeterminate forms, making them a powerful tool when applying rules like l'H么pital's Rule.

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