Problem 35
Question
$$ \lim _{x \rightarrow 0} \frac{e^{x}}{x^{2}}\\{\text { Ans. } \infty\\} $$
Step-by-Step Solution
Verified Answer
As \(x\) approaches \(0\), the denominator \(x^2\) goes to \(0\), while the numerator \(e^x\) goes to \(1\). Since \(e^x\) decreases slower than \(x^2\) for negative \(x\) and increases faster than \(x^2\) for positive \(x\), the ratio \(\frac{e^x}{x^2}\) increases without bound as \(x\) gets closer to \(0\). Thus, the limit of the function is infinite: \( \lim_{x \rightarrow 0} \frac{e^x}{x^2} = \infty \).
1Step 1: Identify the form of the limit
The limit is given by \( \lim_{x \rightarrow 0} \frac{e^x}{x^2} \). We need to find the value of this limit as \( x \) approaches \( 0 \).
2Step 2: Investigate the behavior of the function as x approaches 0
As \(x\) approaches \(0\), we can see that the denominator \(x^2\) goes to \(0\), and the numerator \(e^x\) goes to \(1\), since \(e^0 = 1\). Therefore, the function behaves similar to a \(\frac{1}{0}\) form, but we need to carefully analyze the ratio's behavior.
3Step 3: Compare the rates at which the numerator and denominator approach their respective limits
As \(x\) approaches \(0\), \(e^x\) approaches \(1\) and \(x^2\) approaches \(0\). We have the following behavior:
- \(e^x\) decreases slower than \(x^2\) for negative \(x\).
- \(e^x\) increases faster than \(x^2\) for positive \(x\).
Thus, it seems that the ratio \(\frac{e^x}{x^2}\) would increase without bound for both negative and positive \(x\) approaching \(0\).
4Step 4: Determine the limit
Based on the rate comparison in the previous step, we can conclude that as \(x\) approaches \(0\), the ratio \(\frac{e^x}{x^2}\) increases without bound. Thus, the limit of the function is infinite:
\( \lim_{x \rightarrow 0} \frac{e^x}{x^2} = \infty \).
Key Concepts
Exponential FunctionsBehavior of LimitsInfinite Limits
Exponential Functions
Exponential functions, notably in the form of \( f(x) = e^x \), are a central concept in mathematics because of their unique properties. The base of the exponential function, \( e \), is known as Euler’s number, approximately 2.718. It's a fundamental constant in mathematics.
- Exponential functions grow rapidly as the variable increases.
- They are continuous and differentiable everywhere.
- The derivative of \( e^x \) is easy to work with, as it remains \( e^x \).
Behavior of Limits
The behavior of limits is crucial when evaluating how a function behaves as it approaches a particular value, especially infinity or zero. Limits help us understand the end behavior of functions and their compositions.For example, consider the function \( \frac{e^x}{x^2} \). To understand its limit as \( x \) approaches 0, we study the individual behaviors of the numerator and the denominator:
- The function \( e^x \) approaches 1 as \( x \to 0 \) because \( e^0 = 1 \).
- The denominator \( x^2 \) approaches 0, creating a challenge because dividing by zero typically causes an undefined value.
Infinite Limits
An infinite limit occurs when a function's value grows without bound as the independent variable approaches a certain point, such as in the function \( \frac{e^x}{x^2} \) as \( x \to 0 \). This concept explains situations where limits do not converge to a real number.In our given exercise:
- As the denominator \( x^2 \) approaches 0, it suggests the potential for division by zero.
- The numerator \( e^x \) remains stable, near 1, yet grows to dominate the small values of \( x^2 \).
- This discrepancy causes the function to increase indefinitely, resulting in an infinite limit.
Other exercises in this chapter
Problem 33
$$ \lim _{x \rightarrow \frac{\pi}{2}} \tan x+\tan 2 x \quad\\{\text { Ans. }-\infty,+\infty\\} $$
View solution Problem 34
$$ \lim _{x \rightarrow 0} \cos ^{-1} x \cdot \cot x+\tan ^{-1} x \quad\\{\text { Ans. }+\infty,-\infty\\} $$
View solution Problem 36
$$ \lim _{x \rightarrow \infty} x \ln x\\{\text { Ans. } \infty\\} $$
View solution Problem 37
$$ \lim _{x \rightarrow 1} \frac{\sqrt{x+3}}{\sqrt[3]{x+7}}\\{\text { Ans. } 1\\} $$
View solution