badge image

Enroll For Free Now & Improve Your Performance.

User Icon
User Icon
User Icon
User Icon
User Icon

Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.

Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 12


lim n->infinty (3^n+4^n)^1/n=

9 years ago

Answers : (2)

vikas askiitian expert
509 Points

L = (3n2+4n)1/n  lim n=>infinity

 put n = 1/x

L =   limit  (3/x2+4/x)x  


 can be written as elog(3/x2+4/x)x

L = limit  exlog(3/x2+4/x) x=>0

L = limit e[log(3/x2+4/x) / 1/x]  x=>0

this limit is (infinity/infinity) form so we can use L holpital rule

differentiate numerator & denominator with respect to x & put the limit

L = e0 = 1


9 years ago
Saurabh Max
15 Points

U can do this

take 4^n common n out of bracket

due to^1/n it will become 4

we will have lim [1+(3/4)^n]^1/n*4

now 3/4 is a fraction hence its power infinity will be zero,u can check this from graph 

and whole bracket raised to 1/infinity will be 1 as it is as good as raised to zero

so answer is 4

9 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 101 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details