Question

Limit proofs Use the precise definition of a limit to prove the following limits. Specify a relationship between \(\varepsilon\) and \(\delta\) that guarantees the limit exists. $$\begin{aligned} &\lim _{n \rightarrow 25} \sqrt{x}=5 \text { (Hint: The factorization } x-25=\\\ &(\sqrt{x}-5)(\sqrt{x}+5) \text { implies that } \sqrt{x}-5=\frac{x-25}{\sqrt{x}+5} \end{aligned}$$

Short answer

Question: Prove that the limit exists for the given function using the precise definition of a limit and the provided hint: $$\lim_{x \rightarrow 25} \sqrt{x} = 5.$$ Short Answer: To prove that the limit exists for the given function, we used the precise definition of a limit which states that for any \(\varepsilon > 0\), there exists a \(\delta> 0\) such that for all \(x\) satisfying \(0< |x - 25| < \delta\), we have \(|\sqrt{x} - 5| < \varepsilon\). With the hint given, we rewrote the limit expression as \(\left|\frac{x - 25}{\sqrt{x} + 5}\right| < \varepsilon\), and we found a relationship between \(\varepsilon\) and \(\delta\) by setting \(\delta = 5\varepsilon\). Thus, the limit exists, and the proof is complete.

Step-by-step solution

Write the precise definition of a limit for the given function

According to the precise definition of a limit, we want to show that for any \(\varepsilon > 0\), there exists a \(\delta> 0\) such that for all \(x\) satisfying \(0< |x - 25| < \delta\), we have \(|\sqrt{x} - 5| < \varepsilon\).

Exploit the hint to analyze the limit

We are given the hint that \(x - 25 = (\sqrt{x} - 5)(\sqrt{x} + 5)\) and the suggestion that \(\sqrt{x} - 5 = \frac{x - 25}{\sqrt{x} + 5}\). We can use this identity to rewrite our definition of the limit: $$ \left|\frac{x - 25}{\sqrt{x} + 5}\right| < \varepsilon \quad \text{for } 0 < |x - 25| < \delta. $$

Determine a relationship between \(\varepsilon\) and \(\delta\)

In order to find the relationship between \(\varepsilon\) and \(\delta\), we want to control the absolute value of the expression \(\frac{x - 25}{\sqrt{x} + 5}\) by making it smaller than \(\varepsilon\). To do this, we can take advantage of the fact that \(0 < |x - 25| < \delta\). We have: $$ \left|\frac{x-25}{\sqrt{x}+5}\right| = \frac{|x-25|}{\sqrt{x}+5} < \frac{\delta}{\sqrt{x}+5}. $$ Now, since \(0 < |x - 25| < \delta\), it means that \(0 < |(\sqrt{x} + 5) - 10| < \delta\). In particular, we can suppose that \(|\sqrt{x} + 5 - 10| < \delta\), and to ensure \(\sqrt{x}+5\) is within this interval, we would take \(\delta<5\). Therefore, we have $$ \sqrt{x} + 5 \ge 10 - \delta > 10 - 5 = 5. $$ In this case, $$ \frac{\delta}{\sqrt{x}+5} < \frac{\delta}{5}, $$ which implies $$ \left|\frac{x-25}{\sqrt{x}+5}\right| < \frac{\delta}{5}. $$ So we want this last expression to be smaller than \(\varepsilon\). We can achieve that by making \(\delta = 5\varepsilon\).

Conclusion

We have found a relationship between \(\varepsilon\) and \(\delta\) that satisfies the precise definition of a limit, which guarantees the existence of the limit: $$ \lim_{x \rightarrow 25} \sqrt{x} = 5. $$ For any \(\varepsilon > 0\), we can choose \(\delta = 5\varepsilon\), and the condition for the limit is satisfied.

Key concepts

Epsilon-Delta Definition

The epsilon-delta definition is a core concept in calculus, used to define the limit of a function at a point rigorously. To grasp this, imagine we want to prove \[ \lim_{x \rightarrow a} f(x) = L \]where \( a \) is a point the function \( f(x) \) approaches as \( x \) gets closer but not equal to \( a \), and \( L \) is the limit we're aiming to show.
The definition states that for every small positive number \( \varepsilon \) (epsilon), there exists a small positive number \( \delta \) (delta) such that whenever \( 0 < |x - a| < \delta \), the inequality \( |f(x) - L| < \varepsilon \) holds.
Essentially, \( \varepsilon \) represents how close \( f(x) \) should be to \( L \), and \( \delta \) dictates how close \( x \) needs to be to \( a \) to achieve this closeness. This encapsulates the intuitive idea that by zooming in on the neighborhood of a point \( a \), \( f(x) \) comes arbitrarily close to \( L \).

• \( \varepsilon \) (epsilon): Controls how close the function value \( f(x) \) should be to the limit \( L \).
• \( \delta \) (delta): Controls how close \( x \) needs to be to \( a \), ensuring the function value \( f(x) \) is within \( \varepsilon \) of \( L \).

This is the foundation of proving limits in a precise, mathematical manner.

Sqrt(x) Function

The square root function, denoted as \( \sqrt{x} \), is fundamental and crops up frequently in calculus, especially when dealing with limits and continuous functions. The square root of \( x \) is a number \( y \) such that \( y^2 = x \). It's important to note that the domain of the square root function includes only non-negative numbers, meaning \( x \geq 0 \).
When considering its behavior near a certain point for limits, the square root function is continuous, meaning \( \lim_{x \rightarrow a} \sqrt{x} = \sqrt{a} \) for any \( a \geq 0 \). This continuity becomes handy when evaluating limits as it allows us to apply the limit directly.
For the particular case \( \lim_{x \rightarrow 25} \sqrt{x} = 5 \), we're examining what happens to \( \sqrt{x} \) as \( x \) approaches 25. Since 25 is a "nice number," being a perfect square, we know directly that \( \lim_{x \rightarrow 25} \sqrt{x} = \sqrt{25} = 5 \), confirming the continuity assumption.

• The square root function is continuous for all \( x \geq 0 \).
• \( \sqrt{25} = 5 \), illustrating a straightforward limit case.
• It is essential for understanding limits in calculus and is often involved in factorization techniques.

Factorization

Factorization is the process of breaking down an expression into a product of simpler "factors". In calculus, factorization becomes particularly useful when simplifying limit expressions or rearranging terms to fit into our epsilon-delta framework.
In the given limit exercise, one uses the factorization:
\[ x - 25 = (\sqrt{x} - 5)(\sqrt{x} + 5) \]
to analyze \( \lim_{x \rightarrow 25} \sqrt{x} = 5 \). By substituting the factorized form, it becomes easier to manipulate the expressions, particularly useful in showing the epsilon-delta definition.
Factorization helps tackle seemingly complex limit expressions by simplifying and separating them into multipliable components — often revealing useful relationships such as \( \sqrt{x} - 5 \) equating to \( \frac{x - 25}{\sqrt{x} + 5} \).
Such expressions allow easily managing denominators and understanding behavior approaching specific points.

• Facilitates manipulation and simplification of algebraic expressions.
• Useful for exploring limits and employing the epsilon-delta approach.
• Converts complex expressions into manageable forms suitable for further analysis.

Understanding how factorization fits into evaluating limits can be a powerful tool in your calculus toolkit.