Certainly! Let's evaluate the limit [tex]\( \lim _{x \rightarrow \infty} \frac{x^3+3x^2+1}{x+2} \)[/tex].
To find the limit as [tex]\( x \)[/tex] approaches infinity, we need to analyze the behavior of both the numerator and denominator when [tex]\( x \)[/tex] becomes very large.
1. Numerator: [tex]\( x^3 + 3x^2 + 1 \)[/tex]
- The highest degree term in the numerator is [tex]\( x^3 \)[/tex].
- For large values of [tex]\( x \)[/tex], [tex]\( x^3 \)[/tex] will dominate over [tex]\( 3x^2 \)[/tex] and [tex]\( 1 \)[/tex].
2. Denominator: [tex]\( x + 2 \)[/tex]
- The highest degree term in the denominator is [tex]\( x \)[/tex].
- For large values of [tex]\( x \)[/tex], [tex]\( x \)[/tex] will dominate over [tex]\( 2 \)[/tex].
To simplify the expression, we divide both the numerator and the denominator by [tex]\( x \)[/tex], the highest degree of the denominator.
[tex]\[ \frac{x^3 + 3x^2 + 1}{x + 2} = \frac{x^3/x + 3x^2/x + 1/x}{x/x + 2/x} \][/tex]
This simplifies to:
[tex]\[ = \frac{x^2 + 3x + \frac{1}{x}}{1 + \frac{2}{x}} \][/tex]
Now evaluate the limit as [tex]\( x \)[/tex] approaches infinity:
3. As [tex]\( x \to \infty \)[/tex]:
- The term [tex]\( \frac{1}{x} \)[/tex] approaches [tex]\( 0 \)[/tex].
- The term [tex]\( \frac{2}{x} \)[/tex] also approaches [tex]\( 0 \)[/tex].
So the expression simplifies to:
[tex]\[ \frac{x^2 + 3x + 0}{1 + 0} = \frac{x^2 + 3x}{1} = x^2 + 3x \][/tex]
As [tex]\( x \to \infty \)[/tex], the term [tex]\( x^2 \)[/tex] (the highest degree term in the numerator) will dominate over [tex]\( 3x \)[/tex].
Thus, the limit is:
[tex]\[ \boxed{\infty} \][/tex]
This indicates that the result of the limit as [tex]\( x \)[/tex] approaches infinity is infinity.
