Question and Answers Forum

All Questions      Topic List

Limits Questions

Previous in All Question      Next in All Question      

Previous in Limits      Next in Limits      

Question Number 121576 by oustmuchiya@gmail.com last updated on 09/Nov/20

Answered by TANMAY PANACEA last updated on 09/Nov/20

f(x)=((((√(x^2 +15)) −5)((√(x^2 +15)) +5))/((x−2)((√(x^2 +15)) +5)))  =((x^2 +15−25)/((x−2)((√(x^2 +15)) +5)))=((x^2 −10)/((x−2)((√(x^2 +15)) +5)))  at x=2 the function[discontinous  lim_(x→2) f(x)=∞

f(x)=(x2+155)(x2+15+5)(x2)(x2+15+5)=x2+1525(x2)(x2+15+5)=x210(x2)(x2+15+5)atx=2thefunction[discontinouslimfx2(x)=

Answered by TANMAY PANACEA last updated on 09/Nov/20

f(x)=((x(√(1+((15)/x^2 ))) −5)/(x−2))=(((√(1+((15)/x^2 ))) −(5/x))/(1−(2/x)))  so lim_(x→∞)  f(x)  =(((√(1+0)) −0)/(1−0))=1

f(x)=x1+15x25x2=1+15x25x12xsolimxf(x)=1+0010=1

Commented by oustmuchiya@gmail.com last updated on 12/Nov/20

many thanks indeed

manythanksindeed

Terms of Service

Privacy Policy

Contact: info@tinkutara.com