How do you determine the limit of #5/(x²+8x+15) # as x approaches -3?

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Answer 1 · 2016-07-07T09:27:27

#lim_{x to -3^-} 5/(x²+8x+15) = -oo#

#lim_{x to -3^+} 5/(x²+8x+15) = +oo#

#lim_{x to -3} 5/(x²+8x+15)#

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

#= lim_{h to 0} 5/((-3 + h)^2+8(-3 + h)+15)#

#= lim_{h to 0} 5/(9 - 6h + h^2 - 24 + 8h + 15)#

#= lim_{h to 0} 5/( h^2 + 2h)#

#= lim_{h to 0} 5/( 2h)# because #abs( h^2) "<<" abs h#

on the left side of limit, h < 0 so

#lim_{x to -3^-} 5/(x²+8x+15) = -oo#

on the right side, h > 0 so

#lim_{x to -3^+} 5/(x²+8x+15) = +oo#