Answer to Question #103938 in Calculus for BIVEK SAH

Question

Justify that


lim 1/(5-x)^2
x tend to5 = ∞

Answer 1

$\lim\limits _{x\mapsto 5}\frac{1}{(5-x)^2}=\infty\\ x_0=5, f(x)=\frac{1}{(5-x)^2}.$

It must be shown that

$\forall k>0\quad \exist\delta(k)>0, \quad |x-x_0|<\delta$

performed

$|f(x)|>k.$

Consider

$|x-5|<\delta$

then

$\frac{1}{|x-5|}>\frac{1}{\delta}$

because

$\delta>0, \quad |x-5|>0.$

Consider

$|f(x)|=\left|\frac{1}{(5-x)^2}\right|=\frac{1}{|(x-5)^2|}=\\ =\frac{1}{|x-5|^2}>\frac{1}{\delta^2}>k\implies\\ \frac{1}{\delta^2}>k \implies \delta^2<\frac{1}{k} \implies\\ \delta<\frac{1}{\sqrt{k}}.$

So at $\delta<\frac{1}{\sqrt{k}}$ have $\left|\frac{1}{(5-x)^2}\right|>k$ ,

which means that

$\lim\limits _{x\mapsto 5}\frac{1}{(5-x)^2}=\infty.$