Problem 26
Question
Find the indicated limit. \(\lim _{x \rightarrow-2}-3 x\)
Step-by-Step Solution
Verified Answer
The limit as \(x \rightarrow -2\) of the function \(f(x) = -3x\) is \(6\).
1Step 1: Identify the function and value of x
The function is \(f(x) = -3x\), and we need to find the limit as \(x \rightarrow -2\).
2Step 2: Determine if the function is continuous at x = -2
Since the function is linear, it is continuous everywhere, which means it is continuous at x = -2.
3Step 3: Find the limit by substituting x = -2
As the function is continuous at x = -2, we can directly substitute the value of x into the function to find the limit. So,
\[ \lim_{x \rightarrow -2} -3x = -3(-2)\]
4Step 4: Evaluate the expression
After substituting x = -2, we get:
\[ -3(-2) = 6\]
5Step 5: State the final answer
The limit as \(x \rightarrow -2\) of the function \(f(x) = -3x\) is \(6\).
Key Concepts
Continuous FunctionLinear FunctionLimit Evaluation
Continuous Function
Let's delve into the concept of a continuous function, which is central to understanding limits. A continuous function, in its simplest form, is a function that allows you to draw it without lifting your pencil from the paper. More precisely, for any point on the function, if you can predict the value of the function at that point without any unexpected jumps or gaps, the function is continuous at that point.
When we say a function is continuous at a certain point, we mean that three conditions are met: the function is defined at that point, the limit of the function exists at that point, and the limit of the function as it approaches the point is equal to the function's value at that point. For instance, the function in our exercise is a linear function, which is inherently continuous everywhere on its domain. This lack of breaks or holes in linear functions is why limits are straightforward to evaluate for them, as was demonstrated in the textbook solution.
When we say a function is continuous at a certain point, we mean that three conditions are met: the function is defined at that point, the limit of the function exists at that point, and the limit of the function as it approaches the point is equal to the function's value at that point. For instance, the function in our exercise is a linear function, which is inherently continuous everywhere on its domain. This lack of breaks or holes in linear functions is why limits are straightforward to evaluate for them, as was demonstrated in the textbook solution.
Linear Function
Now, let's focus on the properties of a linear function that make it such a delight for evaluating limits. A linear function is of the form
Remarkably, linear functions have no curves or bends; they are uniform and smooth over their entire domain. Moreover, the rate of change in a linear function is constant, which is another way of saying that the slope is the same no matter where you are on the line. This predictability is what makes linear functions continuous everywhere, paving the way for simpler limit calculations. In our exercise,
f(x) = mx + b, where m is the slope and b is the y-intercept. This kind of function creates a straight line when graphed on a coordinate plane.Remarkably, linear functions have no curves or bends; they are uniform and smooth over their entire domain. Moreover, the rate of change in a linear function is constant, which is another way of saying that the slope is the same no matter where you are on the line. This predictability is what makes linear functions continuous everywhere, paving the way for simpler limit calculations. In our exercise,
f(x) = -3x is linear. Hence, there are no surprises when taking the limit as x approaches any particular value.Limit Evaluation
Evaluating the limit of a function as x approaches a certain value is akin to foretelling the function's behavior near that point. Limit evaluation needs different approaches depending on the type of function and the point of interest. For continuous functions, particularly linear ones like our exercise's example, evaluating a limit is as straightforward as it gets — you simply substitute the value of x into the function.
Understanding limits is essential for delving into calculus, as they form the foundation for defining derivatives and integrals. In instances where functions are not continuous, evaluating limits can become more complex, involving techniques like factoring, rationalizing, or applying special limits such as those involving sin(x)/x as x approaches zero. However, as illustrated by the step-by-step solution, no such complexity arises with our linear function, and the limit as x approaches -2 is readily found to be 6 by direct substitution.
Understanding limits is essential for delving into calculus, as they form the foundation for defining derivatives and integrals. In instances where functions are not continuous, evaluating limits can become more complex, involving techniques like factoring, rationalizing, or applying special limits such as those involving sin(x)/x as x approaches zero. However, as illustrated by the step-by-step solution, no such complexity arises with our linear function, and the limit as x approaches -2 is readily found to be 6 by direct substitution.
Other exercises in this chapter
Problem 26
Find the derivative of the function \(f\) by using the rules of differentiation. \(f(x)=5 x^{4 / 3}-\frac{2}{3} x^{3 / 2}+x^{2}-3 x+1\)
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Let \(f(x)=\frac{1}{x-1}\). a. Find the derivative \(f^{\prime}\) of \(\bar{f}\) b. Find an equation of the tangent line to the curve at the point \(\left(-1,-\
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Find the derivative of each function. \(f(x)=\sqrt{x+1}+\sqrt{x-1}\)
View solution Problem 27
Find the derivative of each function. \(f(x)=\frac{(x+1)\left(x^{2}+1\right)}{x-2}\)
View solution