a. Factor the numerator as a difference of squares:
[tex]\displaystyle\lim_{x\to9}\frac{x-9}{\sqrt x-3}=\lim_{x\to9}\frac{(\sqrt x-3)(\sqrt x+3)}{\sqrt x-3}=\lim_{x\to9}(\sqrt x+3)=6[/tex]
c. As [tex]x\to\infty[/tex], the contribution of the terms of degree less than 2 becomes negligible, which means we can write
[tex]\displaystyle\lim_{x\to\infty}\frac{4x^2-4x-8}{x^2-9}=\lim_{x\to\infty}\frac{4x^2}{x^2}=\lim_{x\to\infty}4=4[/tex]
e. Let's first rewrite the root terms with rational exponents:
[tex]\displaystyle\lim_{x\to1}\frac{\sqrt[3]x-x}{\sqrt x-x}=\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}[/tex]
Next we rationalize the numerator and denominator. We do so by recalling
[tex](a-b)(a+b)=a^2-b^2[/tex]
[tex](a-b)(a^2+ab+b^2)=a^3-b^3[/tex]
In particular,
[tex](x^{1/3}-x)(x^{2/3}+x^{4/3}+x^2)=x-x^3[/tex]
[tex](x^{1/2}-x)(x^{1/2}+x)=x-x^2[/tex]
so we have
[tex]\displaystyle\lim_{x\to1}\frac{x^{1/3}-x}{x^{1/2}-x}\cdot\frac{x^{2/3}+x^{4/3}+x^2}{x^{2/3}+x^{4/3}+x^2}\cdot\frac{x^{1/2}+x}{x^{1/2}+x}=\lim_{x\to1}\frac{x-x^3}{x-x^2}\cdot\frac{x^{1/2}+x}{x^{2/3}+x^{4/3}+x^2}[/tex]
For [tex]x\neq0[/tex] and [tex]x\neq1[/tex], we can simplify the first term:
[tex]\dfrac{x-x^3}{x-x^2}=\dfrac{x(1-x^2)}{x(1-x)}=\dfrac{x(1-x)(1+x)}{x(1-x)}=1+x[/tex]
So our limit becomes
[tex]\displaystyle\lim_{x\to1}\frac{(1+x)(x^{1/2}+x)}{x^{2/3}+x^{4/3}+x^2}=\frac{(1+1)(1+1)}{1+1+1}=\frac43[/tex]
