wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let f:0,2R be defined as fx=log21+tanπx4. Then,limn2nf1n+f2n..............f1 is equal to ______


Open in App
Solution

Step 1: Simplifying the limit expression:

Find the value of limn2nf1n+f2n..............f1

Given data

f:0,2R

fx=log21+tanπx4

Consider the given Equation as,

I=limn2nf1n+f2n..............f1I=limn2nf1n+f2n..............fnnI=limn2nr=1nfrnI=limn2nr=1nlog21+tanπ4×rn

Step 2: Forming the Integral function:

Let us assume that

rn=t1ndr=dtWhenr=1,t=or=n,t=1

Then the integral becomes,

I=201log21+tanπ4tdt

Let us assume that

π4=xπ4dt=dxdt=4πdxWhent=0,x=ot=1x=π4

Then the integral becomes,

I=2×4π0π4log21+tanndn......(1)

We know that

abfxdx=abfa+b-xdx

Then the integral becomes,

I=8π0π4log21+tan0+π4-xdxI=8π0π4log21+tanπ4-xdx

We know that

tanA+B=tanA+tanB1+tanAtanBtanπ4=1

Then the integral becomes,

I=8π0π4log21+1-tanx1+tanxdxI=8π0π4log221+tanxdx

We know that

logmn=logm-logn

Then the integral becomes,

I=8π0π4log22-log21+tanxdxI=8π0π4log22dx-8π0π4log21+tanxdx......(2)

Step 3: Adding the equations (1) and (2)

2I=8π0π41.dx-8π0π4log21+tanxdx+8π0π4log21+tanxdx2I=8π0π41.dx2I=8πx0π42I=8π×π42I=2I=1

Hence, ,limn2nf1n+f2n..............f1 is equal to 1.


flag
Suggest Corrections
thumbs-up
7
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Indeterminant Forms
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon