Problem 44
Question
Find the limit. $$\lim _{x \rightarrow-\infty} \frac{|x|}{|x|+1}$$
Step-by-Step Solution
Verified Answer
Answer: The limit of the function as x approaches negative infinity is 0.
1Step 1: Understand the absolute value
In this step, we need to understand how the absolute value function works. The absolute value of a number is always positive or zero. So, if x is a positive number, then |x| = x. If x is a negative number, then |x| = -x.
For the given limit, since x approaches negative infinity, we know that x is a negative number. In this case, |x| = -x.
2Step 2: Substitute the absolute value
Now we can substitute the absolute value of x with -x and rewrite the function as:
$$\frac{-x}{-x+1}$$
3Step 3: Simplify the function
To simplify the function, we can multiply the numerator and the denominator by (-1) to get rid of the negative signs:
$$\frac{-x(-1)}{(-x+1)(-1)}$$
Which gives us:
$$\frac{x}{x-1}$$
4Step 4: Finding the limit
Now, finding the limit is simple. As x approaches negative infinity, the denominator x - 1 will also approach negative infinity making the entire fraction approaching zero.
Therefore, the limit of the function is:
$$\lim _{x \rightarrow-\infty} \frac{x}{x-1} = 0$$
Key Concepts
Absolute Value in CalculusSimplifying Expressions with SubstitutionsInfinity in Calculus
Absolute Value in Calculus
Absolute value is a way to express the distance of a number from zero on the number line, always as a non-negative value. Understanding absolute value is crucial in calculus, especially when dealing with limits involving negative or positive boundaries.
The expression \(|x|\) represents the absolute value of \(x\).
The expression \(|x|\) represents the absolute value of \(x\).
- If \(x\) is positive, \(|x| = x\).
- If \(x\) is negative, \(|x| = -x\).
Simplifying Expressions with Substitutions
Simplifying algebraic expressions is an art in calculus that often involves substitution of complex parts with simpler forms. The ultimate goal is to make the problem easier to solve.
In this exercise, after acknowledging that \(|x| = -x\) for negative \(x\), the original expression \(rac{|x|}{|x|+1}\) becomes \(rac{-x}{-x+1}\).
In this exercise, after acknowledging that \(|x| = -x\) for negative \(x\), the original expression \(rac{|x|}{|x|+1}\) becomes \(rac{-x}{-x+1}\).
- The next crucial step is to eliminate the negative signs for easier manipulation.
- This is done by multiplying both the numerator and the denominator by \(-1\), resulting in \(rac{x}{x-1}\).
Infinity in Calculus
Infinity is a concept that we encounter frequently in calculus. It's crucial to understand how functions behave as their inputs grow indefinitely large or small. This understanding enables us to define limits that provide valuable insights into the nature of the functions and equations.
In the problem given, we're exploring what happens to the expression \(rac{x}{x-1}\) as \(x\) tends to negative infinity.
In the problem given, we're exploring what happens to the expression \(rac{x}{x-1}\) as \(x\) tends to negative infinity.
- As \(x\) grows more negative, both \(x\) and \(x-1\) become very large in magnitude and negative.
- This results in the fractions’ behavior changing, resembling that of the function \(y = 1\), meaning it approaches zero.
- The approach to zero is because \(x-1\) becomes almost equal to \(x\) as both approach negative infinity.
Other exercises in this chapter
Problem 43
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Use numerical or graphical means to find the limit, if it exists. If the limit of f as x approaches c does exist, answer this question: Is it equal to \(f(c) ?\
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Find $$\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$$ $$f(x)=\sqrt{x}$$
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