Problem 28
Question
Let \(f: R \rightarrow[0, \infty)\) be such that \(\lim _{x \rightarrow 5} f(x)\) exists and \(\lim _{x \rightarrow 5} \frac{(f(x))^{2}-9}{\sqrt{|x-5|}}=0\) Then \(\lim _{x \rightarrow 5} f(x)\) equals: (a) 0 (b) 1 (c) 2 (d) 3
Step-by-Step Solution
Verified Answer
The limit \(\lim_{x \rightarrow 5} f(x)\) equals 3.
1Step 1: Understand the Given Limit
We are given that \( \lim_{x \rightarrow 5} \frac{(f(x))^{2}-9}{\sqrt{|x-5|}} = 0 \). This indicates that as \( x \) approaches 5, the expression \((f(x))^2 - 9\) must approach 0 faster than \(\sqrt{|x-5|}\).
2Step 2: Interpret the Expression
The expression \( (f(x))^2 - 9 \) can be rewritten as \( (f(x)-3)(f(x)+3) \). For the limit of the whole expression to be 0, \( f(x)-3 \) must approach 0, meaning \( f(x) \) must approach 3 as \( x \rightarrow 5 \).
3Step 3: Conclusion from Expression Behavior
Since \((f(x))^2 - 9\) approaches 0 as \(x\) approaches 5, it implies \((f(x) - 3)(f(x) + 3)\) approaches 0. This means that either \(f(x) - 3\) or \(f(x) + 3\) approaches 0, and since \(f(x)\) is approaching a positive value, \(f(x) - 3 = 0\).
4Step 4: Calculate the Limit
From \(f(x) - 3 = 0\), we conclude that \(f(x)\) approaches 3 as \(x \rightarrow 5\). Thus, \(\lim_{x \rightarrow 5} f(x) = 3\).
5Step 5: Verify Answer
Based on our calculations, the answer is consistent with option (d), which is \(\lim_{x \rightarrow 5} f(x) = 3\).
Key Concepts
L'Hopital's RuleContinuityFunction Behavior
L'Hopital's Rule
L'Hopital's Rule is a valuable tool in calculus used to solve indeterminate limits. This rule can be applied when you're dealing with limits that result in forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When you encounter these specific cases, L'Hopital's Rule allows you to find the limit by differentiating the numerator and the denominator separately and then taking the limit again.
- Consider a limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \).
- If the initial substitution results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), the rule can be applied.
- Differentiation: Find the derivatives of \( f(x) \) and \( g(x) \), denoted as \( f'(x) \) and \( g'(x) \).
- Re-evaluate the limit \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \).
Continuity
Continuity is a fundamental concept in calculus that describes a function's behavior as "unbroken" or "smooth." When evaluating limits, understanding continuity can help predict a function's behavior as it approaches a certain point.
A function \( f(x) \) is continuous at a point \( c \) if the following three conditions are met:
A function \( f(x) \) is continuous at a point \( c \) if the following three conditions are met:
- \( f(c) \) is defined.
- \( \lim_{x \to c} f(x) \) exists.
- \( \lim_{x \to c} f(x) = f(c) \).
Function Behavior
Understanding function behavior is about comprehending how a function responds as its inputs change. It's essential for evaluating limits and predicting function values around certain points. In mathematical analysis, this involves:
In our exercise, the function's behavior near \( x = 5 \) was critical. We transformed \( (f(x))^2 - 9 \) into \( (f(x) - 3)(f(x) + 3) \) to decipher that \( f(x) \) must be approaching 3. This understanding of function behavior allowed us to confidently solve for \( \lim_{x \to 5} f(x) = 3 \). Recognizing these patterns and expressions is a skill that helps reveal how functions behave, assisting in limit evaluations like the one discussed. In essence, grasping how small changes in \( x \) affect \( f(x) \) is pivotal in calculus.
- Analyzing expressions such as \( (f(x))^2 - 9 \) to understand output changes as \( x \) nears a certain value.
- Recognizing patterns to deduce what \( f(x) \) approaches.
In our exercise, the function's behavior near \( x = 5 \) was critical. We transformed \( (f(x))^2 - 9 \) into \( (f(x) - 3)(f(x) + 3) \) to decipher that \( f(x) \) must be approaching 3. This understanding of function behavior allowed us to confidently solve for \( \lim_{x \to 5} f(x) = 3 \). Recognizing these patterns and expressions is a skill that helps reveal how functions behave, assisting in limit evaluations like the one discussed. In essence, grasping how small changes in \( x \) affect \( f(x) \) is pivotal in calculus.
Other exercises in this chapter
Problem 26
\(\lim _{x \rightarrow 0} \frac{\sin \left(\pi \cos ^{2} x\right)}{x^{2}}\) equals \(\quad\) [Online May 26, 2012] (a) \(-\pi\) (b) 1 (c) \(-1\) (d) \(\pi\)
View solution Problem 27
\(\lim _{x \rightarrow 0}\left(\frac{x-\sin x}{x}\right) \sin \left(\frac{1}{x}\right)\) [Online May 7, 2012] (a) equals 1 (b) equals 0 (c) does not exist (d) e
View solution Problem 29
\(\lim _{x \rightarrow 2}\left(\frac{\sqrt{1-\cos \\{2(x-2)\\}}}{x-2}\right)\) (a) equals \(\sqrt{2}\) (b) equals \(-\sqrt{2}\) (c) equals \(\frac{1}{\sqrt{2}}\
View solution Problem 30
Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be a positive increasing function with \(\lim _{x \rightarrow \infty} \frac{f(3 x)}{f(x)}=1\) then \(\lim
View solution