Evaluate the limit as [tex]\( t \)[/tex] approaches infinity:

[tex]\[ \lim _{t \rightarrow \infty} \frac{t^3 - 2t^2 + 1}{2t^3 + 3t^2 + 5} \][/tex]



Answer :

To find the limit [tex]\(\lim_{t \rightarrow \infty} \frac{t^3 - 2t^2 + 1}{2t^3 + 3t^2 + 5}\)[/tex], we can follow a systematic approach to simplify and evaluate the expression as [tex]\(t\)[/tex] approaches infinity.

1. Identify the highest power of [tex]\(t\)[/tex] in the denominator: The highest power in the denominator [tex]\(2t^3 + 3t^2 + 5\)[/tex] is [tex]\(t^3\)[/tex].

2. Divide both the numerator and denominator by [tex]\(t^3\)[/tex]:
[tex]\[ \frac{t^3 - 2t^2 + 1}{2t^3 + 3t^2 + 5} = \frac{\frac{t^3}{t^3} - \frac{2t^2}{t^3} + \frac{1}{t^3}}{\frac{2t^3}{t^3} + \frac{3t^2}{t^3} + \frac{5}{t^3}} \][/tex]

3. Simplify each term:
[tex]\[ \frac{t^3 - 2t^2 + 1}{2t^3 + 3t^2 + 5} = \frac{1 - \frac{2}{t} + \frac{1}{t^3}}{2 + \frac{3}{t} + \frac{5}{t^3}} \][/tex]

4. Analyze the behavior as [tex]\(t \rightarrow \infty\)[/tex]:
- As [tex]\(t\)[/tex] approaches infinity, [tex]\(\frac{2}{t}\)[/tex] approaches 0.
- As [tex]\(t\)[/tex] approaches infinity, [tex]\(\frac{1}{t^3}\)[/tex] also approaches 0.
- Similarly, [tex]\(\frac{3}{t}\)[/tex] and [tex]\(\frac{5}{t^3}\)[/tex] both approach 0 as well.

5. Substitute these limits into the simplified expression:
[tex]\[ \lim_{t \rightarrow \infty} \frac{1 - \frac{2}{t} + \frac{1}{t^3}}{2 + \frac{3}{t} + \frac{5}{t^3}} = \frac{1 - 0 + 0}{2 + 0 + 0} = \frac{1}{2} \][/tex]

Thus, the limit is:
[tex]\[ \lim_{t \rightarrow \infty} \frac{t^3 - 2t^2 + 1}{2t^3 + 3t^2 + 5} = \frac{1}{2} \][/tex]

So the final answer is:
[tex]\[ \boxed{\frac{1}{2}} \][/tex]