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Chapter 12: Problem 63

In Exercises 49-68, find the limit by direct substitution. $$ \lim_{x \to 3}\ e^x$$

Short Answer

Expert verified
The limit, \(\lim_{x \to 3} e^x\), by direct substitution method is approximately 20.08554.

Step by step solution

01

Identify the limit

In this problem the limit is given as \(\lim_{x \to 3} e^x\). This means we need to find the value of the function \(e^x\) as \(x\) approaches 3.
02

Direct substitution

As the function \(e^x\) is defined and continuous for all \(x\), we can use the direct substitution method. We substitute \(x\) with 3 in the given expression. So, \(e^x\) becomes \(e^3\).
03

Calculate the value

We calculate the value of \(e^3\) to find the limit. Using the known mathematical constant \(e\) approximately equals to 2.71828. Thus \(e^3\) is approximately equal to 20.08554.

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Key Concepts

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

Direct Substitution
Direct substitution is a straightforward technique used in calculus when finding limits. It involves directly replacing the variable in a function with the value it's approaching.
This method works best when the function is continuous at the point of interest. In order to find \(\lim_{x \to a} f(x)\), you simply substitute \(x = a\) into \(f(x)\).
Here are some points to remember:
  • Ensure the function is defined at the point you are substituting.
  • If the function is continuous, direct substitution will yield the exact limit.
  • This method simplifies the process, especially for polynomial, exponential, and trigonometric functions.
Direct substitution for the expression \(\lim_{x \to 3} e^x\) means we substitute \(x = 3\) into the expression to get \(e^3\). This substitution is possible because \(e^x\) is continuous and defined for all real numbers.
Continuous Functions
Continuous functions are essential in calculus, especially for evaluating limits using direct substitution. A function is continuous at a point if the following conditions are met:
  • The function is defined at the point.
  • The limit of the function as it approaches the point is well-defined.
  • The value of the function at that point is equal to the limit value.
Continuous functions have no breaks, jumps, or holes in their graphs, making them smooth and predictable.
For the function \(e^x\), it is continuous for all real numbers, meaning it's defined everywhere, and we can use direct substitution to find limits. When finding \(\lim_{x \to 3} e^x\), the continuous nature of \(e^x\) allows us to calculate it straightforwardly by evaluating \(e^3\). This property greatly simplifies limit evaluation, as there are no interruptions or undefined points to consider.
Exponential Functions
Exponential functions are a key concept in mathematics, often represented in the form \(e^x\), where \(e\) is the base of the natural logarithm, approximately equal to 2.71828.
These functions have several characteristic properties:
  • They are defined for all real numbers \(x\).
  • Their graphs are always positive and increasing.
  • They grow rapidly, which is useful for modeling real-world scenarios like population growth.
The expression \(e^x\) is continuous, meaning it has no breaks or undefined points, which is why we can easily find limits with direct substitution.
For the limit \(\lim_{x \to 3} e^x\), direct substitution is used to find \(e^3\). The exponential nature of \(e^x\) means this process straightforwardly leads to the solution, \(e^3 \approx 20.08554\), by calculating the exponential expression without complex manipulations.

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Most popular questions from this chapter

In Exercises 9-28, find the limit (if it exists). If the limit does not exist, explain why. Use a graphing utility to verify your result graphically. \\[\lim_{x\to \infty} \left(\dfrac{1+5x}{1-4x} \right) \\]

In Exercises 9-28, find the limit (if it exists). If the limit does not exist, explain why. Use a graphing utility to verify your result graphically. \\[\lim_{x\to -\infty} \dfrac{2x^2 - 5x - 12}{1 - 6x - 8x^2} \\]

In Exercises 37-48, use the limit process to find the area of the region between the graph of the function and the x-axis over the specified interval. $$ f(x) = 4x + 1 $$ Interval \( [0, 1] \)

In Exercises 9-28, find the limit (if it exists). If the limit does not exist, explain why. Use a graphing utility to verify your result graphically. \\[\lim_{t\to \infty} \dfrac{4t^2 - 2t + 1}{-3t^2 + 2t + 2} \\]

In Exercises 37-48, use the limit process to find the area of the region between the graph of the function and the x-axis over the specified interval. $$ f(x) = \frac{1}{4} (x^2 + 4x) $$ Interval \( [1, 4] \)
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