Question

Find the limits of $\frac{(a^2-x^2{)}^{\frac{1}{2}}+(a-3{)}^{\frac{3}{2}}}{(a^3-x^3{)}^{\frac{1}{2}}+(a-x{)}^{\frac{1}{2}}},$ when $x=a$.

Answer 1

To find the limits of $L=\frac{(a^2-x^2{)}^{\frac{1}{2}}+(a-3{)}^{\frac{3}{2}}}{(a^3-x^3{)}^{\frac{1}{2}}+(a-x{)}^{\frac{1}{2}}}$, when $x=a$

Put $x=a-h$ ; then the expression

$L=\underset{h\to 0}{lim}\frac{{\{a^2-(a-h{)}^2\}}^{\frac{1}{2}}+h^{\frac{3}{2}}}{{\{a^3-(a-h{)}^3\}}^{\frac{1}{2}}+h^{\frac{1}{2}}}$

$=\underset{h\to 0}{lim}\frac{{\{a^2-a^2-h^2+2ah\}}^{\frac{1}{2}}+h^{\frac{3}{2}}}{{\{a^3-a^3+h^3+3ah(a-h\left)\right\}}^{\frac{1}{2}}+h^{\frac{1}{2}}}$

$=\underset{h\to 0}{lim}\frac{{\{2ah-h^2\}}^{\frac{1}{2}}+h^{\frac{3}{2}}}{{\left\{3ah\right(a-h)-h^3\}}^{\frac{1}{2}}+h^{\frac{1}{2}}}$

Neglecting the higher powers of $h$, we get

$L=\underset{h\to 0}{lim}\frac{\sqrt{2a}.h^{1/2}}{\sqrt{3a^2}.h^{1/2}+h^{1/2}}$

$=\frac{\sqrt{2a}}{\sqrt{3}a+1}$