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Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow+\infty} \frac{\sqrt{x+1}}{\sqrt{x}} $$

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Identify the form of the expression

The given limit is \( \lim _{x \rightarrow+\infty} \frac{\sqrt{x+1}}{\sqrt{x}} \). First, identify if this expression is in an indeterminate form. By taking the limit as \( x \to \infty \), both the numerator \( \sqrt{x+1} \) and the denominator \( \sqrt{x} \) approach infinity, creating an \( \frac{\infty}{\infty} \) form.
02

Simplify the expression

To simplify the expression, factor out the dominant term from the numerator and the denominator. Both \( \sqrt{x+1} \) and \( \sqrt{x} \) are related square root functions. Factor \( \sqrt{x} \) from the numerator: \( \sqrt{x+1} = \sqrt{x(1 + \frac{1}{x})} = \sqrt{x}\sqrt{1 + \frac{1}{x}} \). Thus, the expression simplifies to \( \frac{\sqrt{x}\sqrt{1 + \frac{1}{x}}}{\sqrt{x}} \).
03

Cancel common terms

Once the expression is factored, simplify by canceling out \( \sqrt{x} \) in the numerator and denominator. This leaves us with \( \sqrt{1 + \frac{1}{x}} \).
04

Evaluate the limit as x approaches infinity

Now, find the limit of the simplified expression as \( x \to \infty \). As \( x \to \infty \), \( \frac{1}{x} \to 0 \). Thus, \( \sqrt{1 + \frac{1}{x}} \to \sqrt{1 + 0} = \sqrt{1} = 1 \). Therefore, \( \lim _{x \rightarrow+\infty} \sqrt{1 + \frac{1}{x}} = 1 \).

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Key Concepts

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

l'Hôpital's Rule
When we encounter limits that resolve into forms that don't allow direct evaluation, we might need some extra tools. Enter **l'Hôpital's Rule**. This is a clever trick that can help us evaluate limits that initially seem unsolvable by conventional means. It focuses on functions that yield a ratio of two quantities, both approaching zero or both heading to infinity, like the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).

**What Does It Do?**
  • If a limit results in one of these indeterminate forms, l'Hôpital's Rule states that the limit of a ratio can be found by taking the derivative of the numerator and the derivative of the denominator separately, and then finding the limit of this new ratio.
  • This can sometimes break down more complex expressions into simpler ones that are more easily evaluated.
However, it's not needed in every situation. For example, if you can simplify the expression through algebraic manipulation and find a direct answer, such as in our problem where we eventually simplified to \( \sqrt{1 + \frac{1}{x}} \), l'Hôpital's Rule becomes unnecessary.
Infinite Limits
Infinite limits occur when variables boost towards incredibly large values. Like rockets rising to space, both numerator and denominator in our context shoot towards infinity. Here’s how it works:

**Understanding the Mechanism**
  • An infinite limit often indicates that one or both parts of an expression approach extremely large numbers. This can seem daunting, but the process often masks simpler patterns or convergences behind these massive values.
  • In many cases, by factoring out dominants or reducing terms, we can boil down these infinite expressions to their core essence. In our example, simplifying \( \sqrt{x+1} \) to \( \sqrt{x}\sqrt{1+\frac{1}{x}} \) allowed us to remove the infinities and focus only on the relevant portion.
This enables us to evaluate the core nature of the limit, discovering hidden simplicities where infinitely large numbers once stood.
Indeterminate Forms
In mathematics, expressions like \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \) are known as **indeterminate forms**, because they do not immediately reveal what the actual limit of the function should be. They are like puzzles we need to solve.

**Recognizing Indeterminate Forms**
  • Indeterminate forms appear when direct substitution in a limit does not instantly give us meaningful answers. They signal the necessity for additional work to resolve the uncertainty.
  • Our example, \( \lim _{x \rightarrow+\infty} \frac{\sqrt{x+1}}{\sqrt{x}} \), appeared initially in the \( \frac{\infty}{\infty} \) form, which demanded manipulation and simplification, rather than a straightforward answer.
The journey from indeterminate forms to true answers invites creativity with algebraic and calculus-based techniques, testing our problem-solving skills and mathematical understanding.

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