/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 78 Find the indicated limit. $$\l... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the indicated limit. $$\lim _{x \rightarrow 0} x\left(1-\frac{1}{x}\right)$$

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Rewrite the equation

Rewrite the equation in the form that allows us to apply L'Hôpital's rule. A recognized indeterminate form for L'Hôpital's rule is either '0/0' or '∞/∞'. Hence, express the equation as \(\frac{1}{\frac{1}{x}}-1\) as the same as \(x\left(1-\frac{1}{x}\right)\). So, the equation becomes \(\lim _{x \rightarrow 0}\left(\frac{1}{\frac{1}{x}}-1\right)\) which upon simplifying delivers \(\lim _{x \rightarrow 0}\left(\frac{x-1}{x}\right)\).
02

Apply L'Hôpital's Rule

Now, it is seen that the new structure of the equation is the type '0/0' which allows the direct application of L'Hôpital's rule. The rule states that if the limit of function takes the forms '0/0' or '∞/∞', it can be calculated by taking the derivative of the numerator and denominator separately and taking their limit as 'x' approaches 0. The derivative of \(x-1\) is 1 and the derivative of \(x\) is also 1, so by taking these limits we get \(\lim _{x \rightarrow 0}\left(\frac{1}{1}\right)\).
03

Calculate the Final limit

Calculating the limit of the function yields: \(\frac{1}{1}\) which simplifies down to a final limit value of 1. So, the original limit \(\lim _{x \rightarrow 0} x\left(1-\frac{1}{x}\right)\) is equal to 1.

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Key Concepts

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

Indeterminate Forms
In calculus, an indeterminate form is an expression involving limits that do not initially yield a definite value. These forms are significant because they indicate where straightforward substitution of limits fails and further analysis is necessary. In the context of L'Hôpital's Rule, the most common indeterminate forms are \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\).

These forms occur frequently when computing limits, especially in cases where direct substitution results in undefined expressions. For example, in the exercise, when \(x\) approaches zero, the expression \(x\left(1-\frac{1}{x}\right)\) leads to an indeterminate form of \(0 \cdot \text{-infinity}\), which is transformed into \(\frac{0}{0}\) through algebraic manipulation.

This transformation is crucial because it allows us to utilize L'Hôpital's Rule to resolve the limit.
Limit Calculation
Calculating limits involves finding the value that a function approaches as the input approaches a particular point. It is a foundational concept in calculus that helps determine continuity, the behavior of functions, and solutions to complex problems.

In the exercise, the task is to find \(\lim _{x \rightarrow 0} x\left(1-\frac{1}{x}\right)\). Initially, a direct substitution for \(x = 0\) would not work as it results in undefined expressions due to division by zero.

To calculate this limit, it's often necessary to manipulate the function algebraically. As seen in our example, this involved rewriting the expression to highlight the indeterminate form so that L'Hôpital's Rule can be applied.
  • First, recognize the indeterminate form and manipulate it if necessary.
  • Use appropriate mathematical theorems like L'Hôpital's Rule to simplify and calculate the limit.
  • Confirm the final result to ensure it addresses the original question.
Knowing how to adjust and examine functions for their limits is key to mastering calculus.
Differentiation
Differentiation, a core technique in calculus, involves finding the derivative of a function. The derivative measures the rate at which a function's value changes as its input changes. It is essential for applying L'Hôpital's Rule when dealing with indeterminate forms.

In the provided solution, differentiation is used to address the \(\frac{0}{0}\) form by calculating the derivatives of the numerator and denominator separately. Let's break down this process:
  • The numerator of the expression \(x-1\) has a derivative of 1.
  • The derivative of the denominator \(x\) is also 1.

By applying these derivatives at \(x = 0\), we end up with the simple expression \(\frac{1}{1}\), resulting in a limit value of 1.

This demonstrates how differentiation aids in resolving limits by rewriting complex expressions in more manageable forms, allowing for straightforward calculation.

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Most popular questions from this chapter

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