Question
In Exercises $35-62,$ use appropriate limit laws and theorems to determine the limit of the sequence or show that it diverges.$$b_{n}=\frac{3-4^{n}}{2+7 \cdot 4^{n}}$$
Step 1
So, we factor out $4^n$ from both the numerator and the denominator. $$ b_{n}=\frac{3-4^{n}}{2+7 \cdot 4^{n}} = \frac{4^n(3/4^n - 1)}{4^n(2/4^n + 7)} $$ Show more…
Show all steps
Your feedback will help us improve your experience
Linh Vu and 98 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 $35-62,$ use appropriate limit laws and theorems to determine the limit of the sequence or show that it diverges. $$ a_{n}=\frac{3-4^{n}}{2+7 \cdot 3^{n}} $$
INFINITE SERIES
Sequences
In Exercises $35-62,$ use appropriate limit laws and theorems to determine the limit of the sequence or show that it diverges. $$ a_{n}=\left(2+\frac{4}{n^{2}}\right)^{1 / 3} $$
In Exercises $35-62,$ use the appropriate limit laws and theorems to determine the limit of the sequence or show that it diverges. $$ a_{n}=\frac{3-4^{n}}{2+7 \cdot 3^{n}} $$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD