Answer
See the explanation
Work Step by Step
The limits \( \lim_{x \rightarrow \pm \infty} k \) and \( \lim_{x \rightarrow \pm \infty} \frac{1}{x} \) can be evaluated as follows:
1. For \( \lim_{x \rightarrow \pm \infty} k \), where \( k \) is a constant, the limit is simply \( k \). This means that as \( x \) approaches positive or negative infinity, the value of the constant \( k \) remains the same.
2. For \( \lim_{x \rightarrow \pm \infty} \frac{1}{x} \), as \( x \) approaches positive or negative infinity, the fraction \( \frac{1}{x} \) tends towards zero. This is because the denominator, \( x \), becomes larger and larger, causing the fraction to approach zero.
These results can be extended to other functions by considering the behavior of the function as \( x \) approaches positive or negative infinity:
- If the function contains only a constant term, the limit as \( x \) approaches infinity will be the value of that constant.
- If the function contains a term with a variable in the denominator, such as \( \frac{1}{x} \), then the limit as \( x \) approaches infinity will be zero.
- If the function contains a term with a variable raised to a power, the limit as \( x \) approaches infinity depends on the degree of the variable. For example, for \( x^n \), where \( n \) is a positive integer, as \( x \) approaches infinity, the function will also approach infinity if \( n \) is odd, and it will approach negative infinity if \( n \) is even.
Examples:
1. \( \lim_{x \rightarrow \infty} (3x^2 + 2x - 1) \) - As \( x \) approaches infinity, the dominant term is \( 3x^2 \), so the limit will be \( \infty \).
2. \( \lim_{x \rightarrow -\infty} \frac{5}{x^3} \) - As \( x \) approaches negative infinity, the fraction \( \frac{5}{x^3} \) tends towards zero, so the limit is \( 0 \).
3. \( \lim_{x \rightarrow \infty} \frac{2x + 1}{x - 3} \) - As \( x \) approaches infinity, the dominant term is \( 2x \) in the numerator and \( x \) in the denominator, so the limit is \( 2 \).
