Problem 24
Question
Evaluate the limits. $$ \lim _{x \rightarrow-\infty} \frac{e^{x}}{1+x} $$
Step-by-Step Solution
Verified Answer
The limit is 0.
1Step 1: Analyze the limit form
We need to evaluate \( \lim_{x \to -\infty} \frac{e^x}{1+x} \). As \(x\) approaches \(-\infty\), the exponential function \(e^x\) approaches zero since the exponential decay makes the numerator very small. Meanwhile, the denominator \(1+x\) becomes more and more negative because \(x\) is getting larger in negative value.
2Step 2: Consider the dominant terms
For the denominator, as \(x\) goes to \(-\infty\), \(x\) itself is the dominant term (since adding 1 is negligible compared to \(x\)). Hence, we can approximate the expression as \( \frac{e^x}{x} \), simplifying \( \frac{e^x}{1 + x} \approx \frac{e^x}{x} \).
3Step 3: Simplify the expression
The form \( \frac{e^x}{x} \) is more straightforward. As \(x\) approaches \(-\infty\), \(e^x\) approaches 0 (since the exponential of very large negative numbers is close to 0), while \(x\) itself continues to approach \(-\infty\), making the fraction a positive number tending to 0.
4Step 4: Conclude the limit value
Given that \( \lim_{x \to -\infty} \frac{e^x}{x} = 0 \), we can conclude that the entire expression \( \lim_{x \to -\infty} \frac{e^x}{1+x} = 0 \) due to the same behavior.
Key Concepts
Exponential FunctionsDominant TermsEvaluating Limits at Infinity
Exponential Functions
Exponential functions are a cornerstone of calculus, playing a crucial role in understanding limits, derivatives, and integrals. At its core, an exponential function is of the form \( f(x) = a^x \), where \(a\) is a positive constant. When talking specifically about the natural exponential function \( e^x \), we refer to a base \( e \), where \( e \approx 2.71828 \), known as Euler's number.
Exponential functions exhibit distinct behaviors based on the sign and magnitude of \(x\). When \(x\) is positive, \( e^x \) grows rapidly, showing exponential growth. However, when \(x\) is negative, as in our exercise where \( x \to -\infty \), \( e^x \) approaches zero, demonstrating exponential decay. This decay happens because raising a number to a large negative power results in a small number, converging towards zero.
These properties make exponential functions vital for analyzing the behavior of expressions at both positive and negative infinities, as they indicate how different terms contribute under specific circumstances.
Exponential functions exhibit distinct behaviors based on the sign and magnitude of \(x\). When \(x\) is positive, \( e^x \) grows rapidly, showing exponential growth. However, when \(x\) is negative, as in our exercise where \( x \to -\infty \), \( e^x \) approaches zero, demonstrating exponential decay. This decay happens because raising a number to a large negative power results in a small number, converging towards zero.
These properties make exponential functions vital for analyzing the behavior of expressions at both positive and negative infinities, as they indicate how different terms contribute under specific circumstances.
Dominant Terms
Understanding dominant terms is essential for evaluating and simplifying limits, particularly when expressions grow large or small towards infinity. In our limit \( \lim_{x \to -\infty} \frac{e^x}{1+x} \), assessing dominant terms helps simplify the analysis.
As \(x\) tends towards \(-\infty\), in the expression \(1+x\), the value of \(x\) becomes significantly larger in magnitude compared to 1, albeit negatively. Therefore, \(x\) is the dominant term in the denominator. This allows the simplification \( \frac{e^x}{1+x} \approx \frac{e^x}{x} \), focusing on the dominant component and making it easier to evaluate the limit.
The concept of dominant terms is a technique rooted in comparing the relative growth of different parts of an expression. It identifies which components have the most significant impact on the overall behavior, allowing for approximate but valid simplifications when evaluating limits.
As \(x\) tends towards \(-\infty\), in the expression \(1+x\), the value of \(x\) becomes significantly larger in magnitude compared to 1, albeit negatively. Therefore, \(x\) is the dominant term in the denominator. This allows the simplification \( \frac{e^x}{1+x} \approx \frac{e^x}{x} \), focusing on the dominant component and making it easier to evaluate the limit.
The concept of dominant terms is a technique rooted in comparing the relative growth of different parts of an expression. It identifies which components have the most significant impact on the overall behavior, allowing for approximate but valid simplifications when evaluating limits.
Evaluating Limits at Infinity
Evaluating limits at infinity involves understanding the behavior of functions as their variables approach infinity, either positively or negatively. It's a fundamental aspect of calculus that deals with the asymptotic behavior of functions, determining how they behave as their inputs grow unbounded.
In the given exercise, \( \lim_{x \to -\infty} \frac{e^x}{1+x} \), by simplifying to \( \frac{e^x}{x} \) and analyzing each part as \(x\) approaches \(-\infty\), we show that irrespective of how the denominator grows negatively large, \( e^x \) decays to zero. Therefore, the whole fraction approaches zero.
A successful evaluation usually includes:
In the given exercise, \( \lim_{x \to -\infty} \frac{e^x}{1+x} \), by simplifying to \( \frac{e^x}{x} \) and analyzing each part as \(x\) approaches \(-\infty\), we show that irrespective of how the denominator grows negatively large, \( e^x \) decays to zero. Therefore, the whole fraction approaches zero.
A successful evaluation usually includes:
- Recognizing the behavior of various components of the function, like exponential decay of \( e^x \).
- Simplifying using dominant terms, making complex limits easier to handle.
- Concluding the limit based on the simplified form, confirming whether it converges to a specific value.
Other exercises in this chapter
Problem 23
In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow 1^{-}} \frac{2}{1-x} $$
View solution Problem 24
In Problems 15-24, find the values of \(x \in\) R for which the given functions are both defined and continuous. $$ f(x)=\cos \left(\frac{2 x}{3+x}\right) $$
View solution Problem 24
In Problems 1-32, use a table or a graph to investigate each limit. $$ \lim _{x \rightarrow 2^{+}} \frac{4}{2-x} $$
View solution Problem 25
Let $$ f(x)=\left\\{\begin{array}{ll} x^{2}+2 & \text { for } x \leq 0 \\ x+c & \text { for } x>0 \end{array}\right. $$ (a) Graph \(f(x)\) when \(c=1\), and det
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