Question

Suppose $\underset{x\to 0}{lim}f\left(x\right)=1$ and $\underset{x\to 0}{lim}g\left(x\right)=-5$, then evaluate
$\underset{x\to 0}{lim}\frac{2f\left(x\right)-g\left(x\right)}{\left(f\right(x)+7{)}^{\frac{2}{3}}}$

Answer 1

${lim}_{x\to 0}f\left(x\right)=1$ & ${lim}_{x\to 0}g\left(x\right)=-5$

We know algebra of limits is as

${lim}_{x\to a}(f\left(x\right)\pm g\left(x\right))={lim}_{x\to a}f\left(x\right)\pm {lim}_{x\to a}g\left(x\right)$

$\Rightarrow {lim}_{x\to 0}[f{\left(x\right)}^m.g{\left(x\right)}^m]={\left({lim}_{x\to a}f\left(x\right)\right)}^m.{\left({lim}_{x\to a}g\left(x\right)\right)}^n$

${lim}_{x\to a}kf\left(x\right)=k{lim}_{x\to a}f\left(x\right)$

Now, ${lim}_{x\to 0}\frac{2(f\left(x\right)-g\left(x\right))}{{[f\left(x\right)+7]}^{\frac{2}{3}}}=\frac{2\left(1\right)-(-5)}{{[1+7]}^{\frac{2}{3}}}=\frac{7}{4}$