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Magazine Chapter 10: Problem 29
Find the limit, if it exists. $$ \lim _{x \rightarrow 0^{+}}\left(\frac{1}{\sqrt{x^{2}+1}}-\frac{1}{x}\right) $$
Short Answer
Expert verified
The limit does not exist, as it approaches negative infinity.
Step by step solution
01
Identify the Type of Limit
We need to find \( \lim _{x \rightarrow 0^{+}} \left( \frac{1}{\sqrt{x^{2}+1}} - \frac{1}{x} \right) \). As \( x \) approaches 0 from the positive side, the expression involves subtraction of two fractions. This indicates that we'll need to simplify by finding a common denominator.
02
Simplify and Find a Common Denominator
To simplify \( \frac{1}{\sqrt{x^{2}+1}} - \frac{1}{x} \), we find a common denominator, \( x\sqrt{x^{2}+1} \). The expression becomes:\[\frac{x - \sqrt{x^2 + 1}}{x\sqrt{x^2 + 1}}\]
03
Multiply by the Conjugate
Notice that \( x - \sqrt{x^2 + 1} \) could be simplified by multiplying the numerator and the denominator by the conjugate \( x + \sqrt{x^2 + 1} \). Multiply:\[\frac{(x - \sqrt{x^2 + 1})(x + \sqrt{x^2 + 1})}{x\sqrt{x^2 + 1}(x + \sqrt{x^2 + 1})}\]This simplifies to:\[\frac{x^2 - (x^2 + 1)}{x\sqrt{x^2 + 1}(x + \sqrt{x^2 + 1})} = \frac{-1}{x\sqrt{x^2 + 1}(x + \sqrt{x^2 + 1})}\]
04
Evaluate the Limit
Now evaluate the limit as \( x \rightarrow 0^{+} \):\[\lim_{x \rightarrow 0^{+}} \frac{-1}{x\sqrt{x^2 + 1}(x + \sqrt{x^2 + 1})}\]At \( x = 0 \), the denominator approaches zero and since the numerator is constant (-1), the limit approaches negative infinity. Thus, the limit does not exist.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Evaluation
When we talk about evaluating limits, we're trying to find where a function is heading as the variable approaches a particular point. In this exercise, we need to evaluate the limit \[ \lim_{x \rightarrow 0^{+}} \left(\frac{1}{\sqrt{x^{2}+1}} - \frac{1}{x}\right) \]This notation means we are interested in the behavior of the expression when \(x\) is getting closer and closer to 0 from the positive side. As \(x\) approaches 0 from the right, it's important to examine both fractions \(\frac{1}{\sqrt{x^2+1}}\) and \(\frac{1}{x}\).
- The term \(\frac{1}{x}\) becomes infinitely large as \(x\) gets very small because any number divided by a tiny positive number results in a big positive number.
- By comparison, \(\frac{1}{\sqrt{x^2+1}}\) remains relatively stable since \(\sqrt{x^2+1}\) approaches 1.
Limit Simplification
Simplifying a limit is often the trickiest but the most essential part of limit evaluation. In this specific exercise, our goal is to simplify the expression
- We start by finding a common denominator for the two fractions. Here, it's \(x\sqrt{x^2+1}\).
- The step transforms the expression into: \[ \frac{x - \sqrt{x^2 + 1}}{x\sqrt{x^2 + 1}} \]
- For further simplification, consider multiplying by the conjugate \(x + \sqrt{x^2+1}\). This clever algebraic trick aids in eliminating the square root from our expression ultimately resulting in a simpler form.
One-Sided Limits
One-sided limits are taken specifically from one direction, either from the left or the right of a particular point. In our exercise, we are dealing with the one-sided limit \[\lim_{x \rightarrow 0^{+}} \left(\frac{1}{\sqrt{x^{2}+1}} - \frac{1}{x}\right)\]"0\(^{+}\)" indicates that we're only interested in how the function behaves as \(x\) approaches 0 from the positive side (right).
- One-sided limits help us understand piecewise functions or functions that have different characteristics on different domains.
- For our function, as \(x\) nears zero from the positive side, each term behaves differently.
Infinity Limits
When evaluating infinity limits, we're interested in the value that a function approaches as the variable heads toward infinity or zero, where the function value might become infinitely large or small.
- In our exercise, when evaluating \[ \lim_{x \rightarrow 0^{+}} \frac{-1}{x\sqrt{x^2+1}(x + \sqrt{x^2+1})} \]
- The simplification in the denominator leads its value to approach zero as \(x\) closes in on 0, while the numerator stays fixed at -1.
- This consistent behavior results in the fraction's value moving toward negative infinity. As such, we conclude that the limit doesn't exist in a traditional sense because it endlessly tends to an infinitely negative value.
