# If limit of f(x)=4 as x->c, what the limit of f(x)^(3/2) as x->c?

Apr 3, 2018

$8$.

#### Explanation:

Prerequisite : Suppose that, ${\lim}_{x \to a} F \left(x\right) = l$. If $G$ is

continuous at $l$, then,

${\lim}_{x \to a} \left(G \circ F\right) \left(x\right) = {\lim}_{x \to a} G \left(F \left(x\right)\right) = G \left({\lim}_{x \to a} F \left(x\right)\right) = G \left(l\right)$.

Given that, ${\lim}_{x \to c} f \left(x\right) = 4$.

Now, $G \left(x\right) = {x}^{\frac{3}{2}}$ is continuous on RR, &,

$\left(G \circ f\right) \left(x\right) = G \left(f \left(x\right)\right) = {\left(f \left(x\right)\right)}^{\frac{3}{2}}$.

$\therefore {\lim}_{x \to c} {\left\{f \left(x\right)\right\}}^{\frac{3}{2}}$,

$= {\lim}_{x \to c} \left(G \circ f\right) \left(x\right)$,

$= G \left({\lim}_{x \to c} f \left(x\right)\right)$,

$= G \left(4\right)$.

$\Rightarrow {\lim}_{x \to c} {\left(f \left(x\right)\right)}^{\frac{3}{2}} = {4}^{\frac{3}{2}} = {\left({2}^{2}\right)}^{\frac{3}{2}} = {2}^{3} = 8$.