5.11 Limits of algebraic functions

Problem : Determine limit

$$\underset{x\to 1}{\overset{}{\lim }}\frac{\left(\sqrt{x}-1\right)\left(2x-1\right)}{2x^2-x-1}$$

Solution : Here, indeterminate form is 0/0. We simplify to change indeterminate form and find limit,

$$\Rightarrow \frac{\left(\sqrt{x}-1\right)\left(2x-1\right)}{2x^2-x-1}=\frac{\left(\sqrt{x}-1\right)\left(2x-1\right)}{\left(x-1\right)\left(2x+1\right)}=\frac{\left(2x-1\right)}{\left(\sqrt{x}+1\right)\left(2x+1\right)}$$

This is determinate form. Plugging “1” for x, we have :

$$\Rightarrow L=\frac{1}{6}$$

Problem : Determine limit

$$\underset{x\to 0}{\overset{}{\lim }}\frac{1}{x{\left(8+x\right)}^{\frac{1}{3}}}-\frac{1}{2x}$$

Solution : The indeterminate form is ∞-∞. Simplifying, we have :

$$f\left(x\right)=\frac{2-{\left(8+x\right)}^{\frac{1}{3}}}{2x{\left(8+x\right)}^{\frac{1}{3}}}$$

We know that :

$$a^{1/3}-b^{1/3}=\frac{\left(a-b\right)}{\left(a^{2/3}+a^{1/3}{b}^{1/3}+b^{2/3}\right)}$$

Using this identity :

$$\Rightarrow 2-{\left(8+x\right)}^{\frac{1}{3}}=8^{\frac{1}{3}}-{\left(8+x\right)}^{\frac{1}{3}}$$ $$=\frac{\left(8-8-x\right)}{\left(8^{\frac{2}{3}}+8^{\frac{1}{3}}{8}^{\frac{1}{3}}+8^{\frac{2}{3}}\right)}$$

Substituting in the given expression,

$$=\frac{-x}{2x{\left(8+x\right)}^{\frac{1}{3}}\left(8^{\frac{2}{3}}+8^{\frac{1}{3}}{8}^{\frac{1}{3}}+8^{\frac{2}{3}}\right)}$$ $$=\frac{-1}{2{\left(8+x\right)}^{\frac{1}{3}}\left(8^{\frac{2}{3}}+8^{\frac{1}{3}}{8}^{\frac{1}{3}}+8^{\frac{2}{3}}\right)}$$

This is a determinate form. Plugging “0” for x,

$$L=\frac{-1}{2X{8}^{\frac{1}{3}}\left(8^{\frac{2}{3}}+8^{\frac{1}{3}}{8}^{\frac{1}{3}}+8^{\frac{2}{3}}\right)}=-\frac{1}{48}$$

Problem : Determine limit :

$$\underset{x\to 0}{\overset{}{\lim }}\frac{\sqrt{\left(1-x^2\right)}-\sqrt{\left(1-x\right)}}{x}$$

Solution : Here, indeterminate form is 0/0. We simplify to change indeterminate form and find limit,

$$\frac{\sqrt{\left(1-x^2\right)}-\sqrt{\left(1-x\right)}}{x}=\frac{\left(1-x^2-1+x\right)}{x\{\sqrt{\left(1-x^2\right)}+\sqrt{\left(1-x\right)}\}}=\frac{\left(1-x\right)}{\{\sqrt{\left(1-x^2\right)}+\sqrt{\left(1-x\right)}\}}$$

This simplified form is not indeterminate. Plugging “0” for “x” :

$$L=\frac{1}{2}$$