please solve question no 5

please solve question no 5 Mll- Limits n! Evaluate lim where n! n—.co (n+l)!—n! LEVEL 2 = n x (n — 1) x ...x 2 xl 5. 4+3an , n 2 1, then find the value of lim an If al = 1 and an +1 = 1 lim x sin - x 3+2an limit using L' Hospital Rule

Dear student,
Concept:-  If a sequence ais convergent  to limit a then the subsequence an+1 is also convergent and converges to same limit  a
Assume that limit is convergent 

So we have
​​​​limnan=l=limnan+1Putting in the equation we getl=4+3l3+2l2l2=4l=±2Since a1=1 So that an>0 or l>0Hence l=2Hope that helps Regards

  • 0
What are you looking for?