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