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Answer 1 · 2016-07-01T08:29:45

#lim_(x->25^+)(x^2+1)/(sqrt(x)-5) = oo#

#lim_(x->25^-)(x^2+1)/(sqrt(x)-5) = -oo#

#lim_(x->25)(x^2+1)/(sqrt(x)-5)#

#= lim_(x->25)(x^2+1)/(sqrt(x)-5) *(sqrt(x)+5)/(sqrt(x)+5)#

#= lim_(x->25) ((x^2+1)(sqrt(x)+5))/(x-25)#

if we substitute #x = 25 + epsilon# into the limit [ie #epsilon = x - 25#], with # 0 < abs(epsilon) < < 1#, then it becomes

#= lim_(epsilon->0) (((25+ epsilon)^2+1)(sqrt(25+epsilon)+5))/(epsilon)#

For #epsilon > 0#, this is #+ oo#

so #lim_(x->25^+)(x^2+1)/(sqrt(x)-5) = oo#

and for #epsilon < 0#, this is #- oo#

so #lim_(x->25^-)(x^2+1)/(sqrt(x)-5) = -oo#