Question
-11 Points] DETAILS MY NOTES SCALCET9 2.XP.6.007. Find the limit. (If the limit is infinite, enter ' \( \infty \) ' or ' \( -\infty \) ', as appropriate. If the limit does not otherwise exist, enter DNE \( \lim _{t \rightarrow \infty} \frac{\sqrt{t}+t^{2}}{6 t-t^{2}} \)
Ask by Blake Peters. in the United States
Feb 26,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The limit is -1.
Solution
Calculate the limit \( \lim_{t\rightarrow \infty} \frac{\sqrt{t}+t^{2}}{6t-t^{2}} \).
Evaluate the limit by following steps:
- step0: Evaluate using transformations:
\(\lim _{t\rightarrow +\infty}\left(\frac{\sqrt{t}+t^{2}}{6t-t^{2}}\right)\)
- step1: Rewrite the expression:
\(\lim _{t\rightarrow +\infty}\left(\frac{\left(\sqrt{t^{-3}}+1\right)t^{2}}{\left(\frac{6}{t}-1\right)t^{2}}\right)\)
- step2: Reduce the fraction:
\(\lim _{t\rightarrow +\infty}\left(\frac{\sqrt{t^{-3}}+1}{\frac{6}{t}-1}\right)\)
- step3: Rewrite the expression:
\(\frac{\lim _{t\rightarrow +\infty}\left(\sqrt{t^{-3}}+1\right)}{\lim _{t\rightarrow +\infty}\left(\frac{6}{t}-1\right)}\)
- step4: Calculate:
\(\frac{1}{\lim _{t\rightarrow +\infty}\left(\frac{6}{t}-1\right)}\)
- step5: Calculate:
\(\frac{1}{-1}\)
- step6: Divide the terms:
\(-1\)
The limit of the expression \( \frac{\sqrt{t}+t^{2}}{6t-t^{2}} \) as \( t \) approaches infinity is -1.
Answered by UpStudy AI and reviewed by a Professional Tutor
Like
error msg
Explain 
Simplify this solution Bonus Knowledge
To find the limit \[ \lim _{t \rightarrow \infty} \frac{\sqrt{t}+t^{2}}{6 t-t^{2}}, \] we start by dividing both the numerator and denominator by \(t^2\): \[ \frac{\frac{\sqrt{t}}{t^2} + 1}{\frac{6t}{t^2} - 1} = \frac{\frac{1}{\sqrt{t}} + 1}{\frac{6}{t} - 1}. \] As \(t\) approaches infinity, \(\frac{1}{\sqrt{t}}\) approaches 0 and \(\frac{6}{t}\) also approaches 0. Thus, we simplify this to: \[ \frac{0 + 1}{0 - 1} = \frac{1}{-1} = -1. \] So, \[ \lim _{t \rightarrow \infty} \frac{\sqrt{t}+t^{2}}{6 t-t^{2}} = -1. \] The final answer is -1.
