What is the limit of #(3+h)^-1 -3^(-1/h)# as h approaches 0?

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Answer 1 · 2016-09-10T11:34:27

#lim_(h to 0^-) (3+h)^-1 -3^(-1/h) = -oo#

#lim_(h to 0^+) (3+h)^-1 -3^(-1/h) = 1/3#

so the limit DNE

#lim_(h to 0) (3+h)^-1 -3^(-1/h)#

#= lim_(h to 0) color(blue)(1/(3+h)) - color(red)(1/3^(1/h))#

the limit of the sum is the sum of the limits

#= lim_(h to 0) color(blue)(1/(3+h)) -lim_(h to 0) color(red)(1/3^(1/h))#

#= 1/3 -lim_(h to 0) color(red)(1/3^(1/h))#

the blue term clearly tends to #1/3# for all #0 < abs h " << " 1#, so we are left to consider the red term

#L =- lim_(h to 0) 3^(-1/h)#

#L =- 3^(lim_(h to 0) -1/h)# as the exponential function is continuous

the sign of the exponent will be different coming at this from either side

#- 3^(lim_(h to 0^-) -1/h) = - 3^oo = -oo#

#- 3^(lim_(h to 0^+) -1/h) = - 3^(-oo) = 0#