First slide
Introduction to limits
Question

Match the following lists:

Column -IColumn -II
A. If L=limx1(7x)32(x+1) , then 12L =P. -2
B. If L=limxπ/4tan3xtanxcosx+π4 then -L/4=Q. 2
C. If L=limx1(2x3)(x1)2x2+x3 , then 20L=R. 1

D. If L=limxlogxn[x][x] , where nN 

     ([x] denotes greatest integer less than or equal to x),then -2L=

S. -1

Moderate
Solution

 a. Let x+1=h. Then, 

     limx1(7x)32(x+1)=limh0(8h)1/32h=limh021h81/32h=2limh0113h81h=112

b. we have

     limxπ/4tan3xtanxcos(x+π/4)=limxπ/4tanx(tanx1)(tanx+1)cos(x+π/4)=limxπ/4tanx(sinxcosx)(tanx+1)cosxcos(x+π/4)=limxπ/4tanx(cosxsinx)(tanx+1)cosxcos(x+π/4)=2limxπ/2tanx12cosx12sinx(tanx+1)cosxcos(x+π/4)=2limxπ/4tanx(tanx+1)cosx=2×2×2=4

c.

      limx1(2x3)(x1)2x2+x3=limx1(2x3)(x1)(2x+3)(x1)=limx1(2x3)(x1)(2x+3)(x1)(x+1)=limx1(2x3)(2x+3)(x+1)=23(2+3)(1+1)=1/10

d. 

       limxlogxn[x][x]=limxlogxn[x]limx[x][x]=01=1

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