What is #lim_(xrarr-3) (2x+6)/|x+3|#?

Question

2777 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-07T14:03:20

#lim_(xrarr-3^-) (2x+6)/|x+3| = -2#

#lim_(xrarr-3^+) (2x+6)/|x+3| = +2#

#lim_(xrarr-3) (2x+6)/|x+3|#

let #x = -3 + h, 0 < abs h "<<" 1#

so the lim becomes

#lim_( h to 0) (2(-3+h)+6)/|(-3+h)+3|#

#=lim_( h to 0) (2h)/(|h|) =lim_( h to 0) 2 h/abs h#

for the left sided limit, h < 0 so we have #"-ve"/"+ve" = "-ve"#

so #lim_(xrarr-3^-) (2x+6)/|x+3| = -2#

for the right sided limit, h > 0 so we have #"+ve"/"+ve" = "+ve"#

so #lim_(xrarr-3^+) (2x+6)/|x+3| = +2#