Question

Prove that

lim 1/(x+5)^2 = ∞

x→-5

Answer 1

We should prove that $\forall M > 0 \;\; \exists \delta_M > 0 \;\; \forall x, |x+5| < \delta_M : |f(x)| > M.$

So for M we should determine $\delta_M.$

Let $\dfrac{1}{(x+5)^2} > M \; \Rightarrow \; (x+5)^2 < \dfrac{1}{M}, \\ |x+5| < \dfrac{1}{\sqrt{M}}\;, \delta_M = \dfrac{1}{\sqrt{M}}.$

Therefore, when x tends to -5, the value of function increases and tends to infinity.