Question
In Exercises $63-66,$ find the limit of the sequence using L'HopitalsRule.$$d_{n}=n^{2}\left(\sqrt[3]{n^{3}+1}-n\right)$$
Step 1
We can do this by dividing both the numerator and the denominator by $n^2$: $$ d_{n}=\frac{n^{3}\sqrt[3]{n^{3}+1}-n^{3}}{n^{2}}=\frac{\sqrt[3]{n^{3}+1}-n}{n^{-2}} $$ Show more…
Show all steps
Your feedback will help us improve your experience
Fuzail Shakir and 76 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
In Exercises $63-66,$ find the limit of the sequence using L'Hopitals Rule. $$ c_{n}=n\left(\sqrt{n^{2}+1}-n\right) $$
INFINITE SERIES
Sequences
In Exercises $63-66,$ find the limit of the sequence using L'Hopitals Rule. $$ a_{n}=\frac{(\ln n)^{2}}{n} $$
In Exercises $63-66,$ find the limit of the sequence using L'Hopitals Rule. $$ b_{n}=\sqrt{n} \ln \left(1+\frac{1}{n}\right) $$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD