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In Exercises \(17-24,\) use Theorem 2 to evaluate the limit. $$ \lim _{x \rightarrow 0}\left(x^{2}+2\right) /(x+1) $$

Short Answer

Expert verified
The limit is 2.

Step by step solution

01

Identify Theorem 2

In calculus, Theorem 2 often refers to the limit laws or the Direct Substitution Property. These properties can be used to find limits by substituting the value directly into the function, assuming the function is defined at that point.
02

Apply Direct Substitution

Using the Direct Substitution Property, substitute the value of \( x = 0 \) directly into the function \( \frac{x^2 + 2}{x + 1} \). This gives: \[ \frac{0^2 + 2}{0 + 1} = \frac{2}{1} = 2. \]
03

Verify the Function is Continuous at the Point

Ensure that the function \( f(x) = \frac{x^2 + 2}{x + 1} \) is continuous at \( x = 0 \). This function is continuous for all \( x eq -1 \) since the denominator is non-zero (it is not -1 at \( x = 0 \)). Thus, the limit can be correctly evaluated using direct substitution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direct Substitution Property
In calculus, the Direct Substitution Property is the first thing we should check when evaluating limits. It states that if a function is continuous at a given point, you can substitute the value directly in to find the limit. This method is quick and efficient when applicable.

For example, consider the function \( f(x) = \frac{x^2 + 2}{x + 1} \). To apply direct substitution, simply plug \( x = 0 \) into the function without altering it:
  • Substitute \( 0 \) for \( x \)
  • Calculate \( \frac{0^2 + 2}{0 + 1} = \frac{2}{1} = 2 \)
This process saves time and shows how powerful the Direct Substitution Property can be when the function is nicely behaved.
Continuity of Functions
Continuity plays a crucial role in calculus, especially when dealing with limits. A function is continuous at a point if there are no breaks, jumps, or holes in its graph at that point. In simpler terms, you can draw the curve at that point without lifting your pencil from the paper.

For a function like \( f(x) = \frac{x^2 + 2}{x + 1} \), checking its continuity at \( x = 0 \) involves verifying that the function is defined, and its denominator isn’t zero. Since the denominator \( x + 1 \) equals 1 when \( x = 0 \), the function is indeed continuous at \( x = 0 \). This confirmation allows us to use direct substitution to find limits freely, without running into undefined expressions or infinite results.
Limit Laws
Limit laws are essential rules in calculus used to evaluate limits when functions are not so straightforward. They provide a set of guidelines that allow you to simplify limit expressions.
  • Sum/Difference Law: The limit of a sum or difference is the sum or difference of the limits.
  • Product Law: The limit of a product is the product of the limits.
  • Quotient Law: The limit of a quotient is the quotient of the limits, provided the limit of the denominator is not zero.
  • Power Law: The limit of a power is the power of the limit.
These laws help break down complex limit calculations into simpler parts, making evaluations easier. In our exercise, most of these didn’t need to be explicitly used due to direct substitution being valid. However, they are critical tools for more intricate expressions.
Evaluating Limits
Evaluating limits involves determining the behavior of a function as it approaches a certain point. It’s like predicting where the function is heading as it gets closer to this point.

In our example, evaluating \( \lim _{x \rightarrow 0} \frac{x^2 + 2}{x + 1} \) means predicting what output the function gives when \( x \) is just about zero. When evaluating, consider the following steps:
1. **Apply Direct Substitution:** Try inserting the value directly.2. **Check for Continuity:** Ensure the function is continuous at that point.3. **Use Limit Laws:** If direct substitution doesn’t work, apply limit laws to modify and simplify the expression.Through this careful evaluation, we achieve a more thorough understanding of the function’s behavior. In this problem, direct substitution worked perfectly because we verified that the function is continuous at \( x = 0 \).

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