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Sagot :
To find the horizontal asymptote of the function [tex]\( f(x) = \frac{3x^2 - 5}{x + 1} \)[/tex], we need to analyze its behavior as [tex]\( x \)[/tex] approaches infinity ([tex]\( x \to \infty \)[/tex]).
### Step-by-Step Solution:
1. Identify the function:
[tex]\[ f(x) = \frac{3x^2 - 5}{x + 1} \][/tex]
2. Understand the concept of horizontal asymptotes:
Horizontal asymptotes occur when [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] or [tex]\(-\infty\)[/tex] and provide a value that the function approaches. We are particularly interested in the behavior as [tex]\( x \to \infty \)[/tex].
3. Consider the behavior as [tex]\( x \to \infty \)[/tex]:
To find the horizontal asymptote, we take the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{3x^2 - 5}{x + 1} \][/tex]
4. Simplify the expression within the limit:
When [tex]\( x \to \infty \)[/tex], the highest degree terms in the numerator and denominator dominate the behavior of the function:
[tex]\[ \lim_{x \to \infty} \frac{3x^2 - 5}{x + 1} = \lim_{x \to \infty} \frac{3x^2}{x + 1} \approx \lim_{x \to \infty} \frac{3x^2}{x} \quad (\text{for large } x) \][/tex]
5. Divide numerator and denominator by [tex]\( x \)[/tex]:
Simplify by dividing the numerator and the denominator by [tex]\( x \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{3x^2 / x}{(x + 1) / x} = \lim_{x \to \infty} \frac{3x}{1 + \frac{1}{x}} \][/tex]
6. Evaluate the simplified limit:
As [tex]\( x \to \infty \)[/tex], the term [tex]\(\frac{1}{x}\)[/tex] approaches 0:
[tex]\[ \lim_{x \to \infty} \frac{3x}{1 + \frac{1}{x}} = \lim_{x \to \infty} \frac{3x}{1 + 0} = \lim_{x \to \infty} 3x \][/tex]
Since [tex]\( 3x \)[/tex] grows without bound as [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} 3x = \infty \][/tex]
Thus, the function [tex]\( f(x) = \frac{3x^2 - 5}{x + 1} \)[/tex] does not approach a finite value but instead increases without bound as [tex]\( x \to \infty \)[/tex].
### Conclusion:
The horizontal asymptote is:
[tex]\[ \lim_{x \to \infty} f(x) = \infty \][/tex]
Therefore, [tex]\( f(x) \to \infty \)[/tex] as [tex]\( x \to \infty \)[/tex], indicating that there is no finite horizontal asymptote but rather the function increases indefinitely.
### Step-by-Step Solution:
1. Identify the function:
[tex]\[ f(x) = \frac{3x^2 - 5}{x + 1} \][/tex]
2. Understand the concept of horizontal asymptotes:
Horizontal asymptotes occur when [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex] or [tex]\(-\infty\)[/tex] and provide a value that the function approaches. We are particularly interested in the behavior as [tex]\( x \to \infty \)[/tex].
3. Consider the behavior as [tex]\( x \to \infty \)[/tex]:
To find the horizontal asymptote, we take the limit of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\( \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{3x^2 - 5}{x + 1} \][/tex]
4. Simplify the expression within the limit:
When [tex]\( x \to \infty \)[/tex], the highest degree terms in the numerator and denominator dominate the behavior of the function:
[tex]\[ \lim_{x \to \infty} \frac{3x^2 - 5}{x + 1} = \lim_{x \to \infty} \frac{3x^2}{x + 1} \approx \lim_{x \to \infty} \frac{3x^2}{x} \quad (\text{for large } x) \][/tex]
5. Divide numerator and denominator by [tex]\( x \)[/tex]:
Simplify by dividing the numerator and the denominator by [tex]\( x \)[/tex]:
[tex]\[ \lim_{x \to \infty} \frac{3x^2 / x}{(x + 1) / x} = \lim_{x \to \infty} \frac{3x}{1 + \frac{1}{x}} \][/tex]
6. Evaluate the simplified limit:
As [tex]\( x \to \infty \)[/tex], the term [tex]\(\frac{1}{x}\)[/tex] approaches 0:
[tex]\[ \lim_{x \to \infty} \frac{3x}{1 + \frac{1}{x}} = \lim_{x \to \infty} \frac{3x}{1 + 0} = \lim_{x \to \infty} 3x \][/tex]
Since [tex]\( 3x \)[/tex] grows without bound as [tex]\( x \to \infty \)[/tex]:
[tex]\[ \lim_{x \to \infty} 3x = \infty \][/tex]
Thus, the function [tex]\( f(x) = \frac{3x^2 - 5}{x + 1} \)[/tex] does not approach a finite value but instead increases without bound as [tex]\( x \to \infty \)[/tex].
### Conclusion:
The horizontal asymptote is:
[tex]\[ \lim_{x \to \infty} f(x) = \infty \][/tex]
Therefore, [tex]\( f(x) \to \infty \)[/tex] as [tex]\( x \to \infty \)[/tex], indicating that there is no finite horizontal asymptote but rather the function increases indefinitely.
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