Calculator limits Use a calculator to approximate the following limits. lim x🠂0 e^3x-1 / x
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Step 1: Recognize that the limit \( \lim_{x \to 0} \frac{e^{3x} - 1}{x} \) is an indeterminate form of type \( \frac{0}{0} \). This suggests that L'Hôpital's Rule might be applicable.
Step 2: Recall L'Hôpital's Rule, which states that if \( \lim_{x \to c} \frac{f(x)}{g(x)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), provided the limit on the right exists.
Step 3: Differentiate the numerator and the denominator separately. The derivative of the numerator \( e^{3x} - 1 \) is \( 3e^{3x} \), and the derivative of the denominator \( x \) is \( 1 \).
Step 4: Apply L'Hôpital's Rule to the original limit: \( \lim_{x \to 0} \frac{e^{3x} - 1}{x} = \lim_{x \to 0} \frac{3e^{3x}}{1} \).
Step 5: Evaluate the new limit: \( \lim_{x \to 0} 3e^{3x} \). As \( x \to 0 \), \( e^{3x} \to e^0 = 1 \). Therefore, the limit simplifies to \( 3 \times 1 = 3 \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Limits
A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a certain value. It helps in understanding how functions behave near points of interest, including points of discontinuity or infinity. In this case, we are interested in the limit of the function as x approaches 0.
Exponential functions are mathematical functions of the form f(x) = a * e^(bx), where e is the base of natural logarithms. They are characterized by their rapid growth or decay and are essential in various applications, including calculus. The expression e^(3x) in the limit indicates that we are dealing with an exponential function, which will influence the limit's value as x approaches 0.
L'Hôpital's Rule is a method used to evaluate limits of indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of f(x)/g(x) results in an indeterminate form, the limit can be found by taking the derivative of the numerator and the derivative of the denominator. This rule is particularly useful in this problem, as substituting x = 0 directly leads to an indeterminate form.