Problem 49
Question
, find the limit or state that it does not exist. $$ \lim _{x \rightarrow 0} \sqrt{|x|} $$
Step-by-Step Solution
Verified Answer
The limit is 0.
1Step 1: Understanding the Problem
We need to find the limit \( \lim_{x \rightarrow 0} \sqrt{|x|} \). This expression involves the square root of the absolute value of \( x \). The absolute value affects both the negative and positive paths towards zero.
2Step 2: Consider Approaching from the Positive Side
First, consider the limit as \( x \to 0^+ \) (approaching zero from the positive side). The expression becomes \( \sqrt{x} \). As \( x \) approaches zero, the square root \( \sqrt{x} \) also approaches zero. Thus, \( \lim_{x \to 0^+} \sqrt{x} = 0 \).
3Step 3: Consider Approaching from the Negative Side
Next, consider the limit as \( x \to 0^- \) (approaching zero from the negative side). Here \( |x| \) equals \(-x\), so the expression is \( \sqrt{-x} = \sqrt{|x|} \). As \( x \) approaches zero from the negative side, \( \sqrt{|x|} = \sqrt{-x} \) also approaches zero. Thus, \( \lim_{x \to 0^-} \sqrt{|x|} = 0 \).
4Step 4: Finalizing the Conclusion
Since both one-sided limits (from 0^+ and 0^-) equal zero, the two-sided limit also equals zero. Therefore, \( \lim_{x \to 0} \sqrt{|x|} = 0 \).
Key Concepts
Absolute ValueContinuous FunctionsOne-sided Limits
Absolute Value
In mathematics, the absolute value of a number is its distance from zero on the number line, regardless of direction. It is denoted by vertical bars, such as \(|x|\). Absolute value is always non-negative. Here are some key points:
- If \(x\) is positive, \(|x| = x\).
- If \(x\) is negative, \(|x| = -x\).
- If \(x\) is zero, \(|x| = 0\).
Continuous Functions
A continuous function is one where small changes in input lead to small changes in output, with no abrupt jumps or breaks. In simple terms, you can draw the graph of a continuous function without lifting your pencil. This continuity ensures that limits can be effectively calculated.
When dealing with continuous functions, particularly when involving an absolute value, the function behaves predictably. In the exercise at hand, \(\sqrt{|x|}\) is continuous everywhere except at \(x = 0\). At this point, we look at the behavior from both sides (as \(x\) approaches 0 from the negative and positive sides). Since \(\sqrt{|x|}\) smoothly transitions through all values, it demonstrates continuity by yielding a consistent limit of 0 from both sides.
When dealing with continuous functions, particularly when involving an absolute value, the function behaves predictably. In the exercise at hand, \(\sqrt{|x|}\) is continuous everywhere except at \(x = 0\). At this point, we look at the behavior from both sides (as \(x\) approaches 0 from the negative and positive sides). Since \(\sqrt{|x|}\) smoothly transitions through all values, it demonstrates continuity by yielding a consistent limit of 0 from both sides.
- Continuous functions simplify the process of finding limits as they behave predictably across their domain.
- Being able to rely on continuity across a function's domain allows for the simplification of calculations.
One-sided Limits
One-sided limits focus on investigating the behavior of a function as it approaches a specific point from one side only - either the left or the right.
- \(\lim_{x \to a^+} f(x)\) refers to the limit of function \(f(x)\) as \(x\) approaches \(a\) from the right (positive direction).
- \(\lim_{x \to a^-} f(x)\) refers to the limit of function \(f(x)\) as \(x\) approaches \(a\) from the left (negative direction).
- When approaching 0 from the positive side, \(x > 0\) and \(|x| = x\), so \(\lim_{x \to 0^+} \sqrt{x} = 0\).
- When approaching 0 from the negative side, \(x < 0\) and \(|x| = -x\), which ensures \(\lim_{x \to 0^-} \sqrt{|x|} = 0\).
Other exercises in this chapter
Problem 49
The line \(y=a x+b\) is called an oblique asymptote to the graph of \(y=f(x)\) if either \(\lim _{x \rightarrow \infty}[f(x)-(a x+b)]=0\) or \(\lim _{x \rightar
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Determine the largest interval over which the given function is continuous. $$ f(x)=\frac{1}{\sqrt{25-x^{2}}} $$
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Verify that the given equations are identities. \(\tanh (x-y)=\frac{\tanh x-\tanh y}{1-\tanh x \tanh y}\)
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