Evaluating Limits. Does the limit of $x ->-5$ exist in the function $(x^3)/(x+5)^2$ ?

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Evaluating Limits. Does the limit of $x ->-5$ exist in the function $(x^3)/(x+5)^2$ ?

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Answer 1 · 2018-04-17T19:18:31

$lim_(x->-5) x^3/(x-5)^2 =-oo$

Explanation:

The numerator is continuous for $x=-5$ as:

$[x^3]_(x=-5) = -125$

The denominator instead vanishes as:

$lim_(x->-5) (x+5)^2 = 0$

and we can note that it approaches $0$ only with positive values as:

$(x+5)^2 > 0$ for $x!= -5$

Around $x=-5$ then the ratio:

$x^3/(x+5)^2 < 0$

is negative and not bounded, so we can conclude that:

$lim_(x->-5) x^3/(x-5)^2 =-oo$

We can formally demonstrate is by choosing any number $M >0$ and then a number $delta_M > 0$ such that $delta_M < min(4, 1/sqrt(M))$ .

Then, for $x in (-5-delta_M, -5+delta_M)$ we have that, as $x^3$ is strictly increasing:

$x^3 <= (-5+delta_M)^3$

and then as $delta_M < 4$

$x^3 <= (-5+4)^3$

that is:

$(1) " "x^3 <= -1$

while as:

$abs (x+5) < delta_M$

$(x+5)^2 < delta_M^2$

and as $delta_M < sqrt(1/M)$ :

$(x+5)^2 < (sqrt(1/M))^2$

that is:

$(x+5)^2 < 1/M$

and taking the reciprocal:

$(2) " " 1/(x+5)^2 > M$

Multiplying the inequalities $(1)$ and $(2)$ we finally have that $x in (-5-delta_M, -5+delta_M)$ implies

$x^3/(x+5)^2 < -M$

which proves that:

$lim_(x->-5) x^3/(x-5)^2 =-oo$

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Evaluating Limits. Does the limit of $x ->-5$ exist in the function $(x^3)/(x+5)^2$ ?