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Magazine Chapter 3: Problem 78
Find the limit by interpreting the expression as an appropriate derivative. $$ \lim _{x \rightarrow 0} \frac{\exp \left(x^{2}\right)-1}{x} $$
Short Answer
Expert verified
The limit is 0.
Step by step solution
01
Recognize the form of the expression
We have the limit \( \lim_{x \rightarrow 0} \frac{\exp(x^2) - 1}{x} \). Recognize that this expression takes the form of \( \frac{f(x) - f(a)}{x-a} \) which resembles the definition of the derivative of a function \( f \) at a point \( a \).
02
Consider the function and point
Observe that the expression inside the limit can be related to the derivative of the function \( f(u) = \exp(u) \) evaluated at \( u = 0 \). Rewrite \( u \) as \( x^2 \) so that the derivative \( f'(u) \) at \( u=0 \) will help us in finding our limit.
03
Differentiate the function
Find the derivative of \( f(u) = \exp(u) \) as \( f'(u) = \exp(u) \). At \( u=0 \), \( f'(0) = \exp(0) = 1 \). This derivative corresponds to the slope of the tangent line of the function \( \exp(u) \) at \( u = 0 \).
04
Apply the Chain Rule
Recognize that we can use the chain rule for derivatives. For \( f(x) = \exp(x^2) \), the derivative \( f'(x) \) is obtained by the chain rule as \( f'(x) = \exp(x^2) \cdot (2x) \).
05
Evaluate the Limit
Since we are looking for \( \lim_{x \rightarrow 0} \frac{\exp(x^2) - 1}{x} \), recognize that \( x^2 \to 0 \) as \( x \to 0 \). Utilize the fact that this form suggests a derivative of \( \exp(u) \) at \( u = 0 \), and so the given expression converges to \( 0 \cdot f'(0) = 2x \cdot 1 = 0 \). Thus, the limit is \( 0 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limit Calculation
Calculating limits is a fundamental concept in calculus. It involves finding the value that a function approaches as the input approaches a certain point. In the exercise, we started with the limit \( \lim_{x \rightarrow 0} \frac{\exp(x^2) - 1}{x} \). To solve such limits, it is useful to recognize forms and apply derivative principles. This specific problem involves interpreting the limit as a derivative. To do this, you should:
- Recognize the given expression in the limit resembles a derivative form, \( \frac{f(x) - f(a)}{x-a} \), at a point \( a \).
- Understand that you can treat it as finding the derivative of a function. This often simplifies the task of limit calculation.
- Apply concepts like L'H么pital's Rule if necessary, when dealing with indeterminate forms.
Derivative
The derivative is a central concept of calculus. It signifies the instantaneous rate of change of a function concerning its variables. In the provided problem, the function we consider is \( f(u) = \exp(u) \). At the core:
- The derivative of a function, \( f(x) \), is denoted \( f'(x) \) or \( \frac{df}{dx} \), representing how \( f \) changes as \( x \) changes.
- To find the derivative of \( \exp(u) \), recall that the derivative remains \( \exp(u) \) since the exponential function's rate of increase is proportional to its value.
- Evaluating at \( u=0 \) gives \( f'(0) = \exp(0) = 1 \), showing the slope of the tangent at this point is 1. This critical value helps in deducing later steps in problems like these.
Chain Rule
The Chain Rule is a method in calculus for computing the derivative of composite functions. When two functions are combined, such as in the expression \( f(x) = \exp(x^2) \), the Chain Rule helps find the derivative with the following steps:
- First, identify the "inner" function and "outer" function. In our problem, \( x^2 \) serves as the inner function and \( \exp(u) \) as the outer function.
- The Chain Rule states that for two functions \( g \) and \( f \), derivative of \( f(g(x)) \) is \( f'(g(x)) \times g'(x) \).
- In our solution, we recognize \( f'(x) = \exp(x^2) \cdot (2x) \). Here, \( \exp(x^2) \) is differentiated as \( \exp(x^2) \), and the derivative of \( x^2 \) is \( 2x \).
