Problem 1 - Calculus Single Variable

Question

In each of Exercises \(1-16,\) use l'Hôpital's Rule to find the limit, if it exists. \(\lim _{x \rightarrow 0}\left(1-e^{x}\right) / x\)

Step-by-Step Solution

Answer

The limit is \(-1\).

Step 1: Identify the Indeterminate Form

To use l'Hôpital's Rule, first check if the limit \[ \lim _{x \rightarrow 0}\left(1-e^{x}\right) / x \]produces an indeterminate form of \( \frac{0}{0} \). Substituting \( x = 0 \), we have \( 1 - e^0 = 0 \) in the numerator and \( x = 0 \) in the denominator, confirming both are zero. Thus, this is an indeterminate form \( \frac{0}{0} \).

Step 2: Apply l'Hôpital's Rule

l'Hôpital's Rule can be applied to indeterminate forms of \( \frac{0}{0} \) by taking the derivative of the numerator and denominator separately.The derivative of the numerator \( 1 - e^x \) is \( -e^x \), and the derivative of the denominator \( x \) is \( 1 \). Using l'Hôpital's Rule, the limit becomes:\[\lim _{x \rightarrow 0} \frac{-e^x}{1}\]

Step 3: Compute the New Limit

Now, evaluate the new limit:\[\lim _{x \rightarrow 0} -e^x\]Substitute \( x = 0 \) into \( -e^x \):\(-e^0 = -1\).Thus, the limit evaluates to \(-1\).

Key Concepts

Indeterminate Form

When trying to evaluate a limit, you might stumble upon expressions that seem undefined. This can be due to an indeterminate form, which is a result that does not provide enough information to determine the limit's value immediately.
One common indeterminate form occurs as \( \frac{0}{0} \). When both the numerator and denominator of a fraction approach zero as \( x \) approaches a certain point, like in the function \( \lim _{x \rightarrow 0}\left(1-e^{x}\right) / x \), it signals an indeterminate form.

• This does not mean that the limit does not exist; rather, it requires more work to solve.
• Recognizing an indeterminate form is the first step before applying other mathematical rules such as l'Hôpital's Rule.

Identifying indeterminate forms alerts you to use specific techniques for evaluating challenging limits.

Limit Evaluation

Limit evaluation is a fundamental concept in calculus used to find the value that a function approaches as the input approaches a certain point. When standard algebraic simplification isn’t enough, you may need more advanced methods. In the case where an indeterminate form is present, l'Hôpital's Rule is a powerful tool.
To evaluate limits using l'Hôpital's Rule:

• Ensure that the function results in an indeterminate form such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Take the derivative of both the numerator and denominator separately.
• Re-evaluate the limit with these new functions.

For example, in \( \lim _{x \rightarrow 0}\left(1-e^{x}\right) / x \), after confirming it is \( \frac{0}{0} \), you differentiate to get \( \lim _{x \rightarrow 0} \frac{-e^x}{1} \), which simplifies to finding \( \lim _{x \rightarrow 0} -e^x = -1 \).
This process simplifies finding what value the function heads towards as \( x \) nears a point.

Derivatives

Derivatives play a crucial role when discussing limit evaluation techniques like l'Hôpital's Rule. Derivatives express the rate at which a function is changing at any point. They are central to transforming complex limits into easier-to-handle forms.
In the context of using l’Hôpital's Rule, here's how derivatives are typically employed:

• Calculating the derivative of the numerator lets us understand how the top part of a fraction behaves near a given point.
• Likewise, the derivative of the denominator helps identify the behavior of the bottom part.

These newly derived functions are then substituted back into the limit to evaluate it more straightforwardly. For example, from \(1-e^x\) we derive \(-e^x\), and from \(x\), we derive \(1\), simplifying the limit evaluation. This derivative-based approach allows us to bypass direct computation issues like \( \frac{0}{0} \), leading to more robust solutions.