Math Algebraically Rationalizing and Negative Exponents Solution

Honors Pre-CalculusLimits D5 WorksheetEvaluating Limits Algebraically Find the limit, if it exists. [lim_{x\rightarrow0}\frac{1-\sqrt{x^{2}+1}}{x^{2}}=\frac{0}{0}] [lim_{x\rightarrow0}\frac{1-\sqrt{x^{2}+1}}{x^{2}}\cdot\frac{1+\sqrt{x^{2}+1}}{1+\sqrt{x^{2}+1}}=] [lim_{x\rightarrow0}\frac{1-(x^{2}+1)}{x^{2}(1+\sqrt{x^{2}+1})}=lim_{x\rightarrow0}\frac{-x^{2}}{x^{2}(1+\sqrt{x^{2}+1})}=] [lim_{x\rightarrow0}\frac{-1}{1+\sqrt{x^{2}+1}}=\boxed{-\frac{1}{2}}] [lim_{x\rightarrow7}\frac{\sqrt{7-x}-\sqrt{7}}{x}=\frac{0}{0}] [lim_{x\rightarrow7}\frac{\sqrt{7-x}-\sqrt{7}}{x}\cdot\frac{\sqrt{7-x}+\sqrt{7}}{\sqrt{7-x}+\sqrt{7}}=] [lim_{x\rightarrow7}\frac{7-x-7}{x(\sqrt{7-x}+\sqrt{7})}=lim_{x\rightarrow7}\frac{-x}{x(\sqrt{7-x}+\sqrt{7})}=] [lim_{x\rightarrow7}\frac{-1}{\sqrt{7-x}+\sqrt{7}}=\boxed{-\frac{\sqrt{7}}{14}}] [lim_{x\rightarrow2}\frac{4-\sqrt{18-x}}{x-2}=\frac{0}{0}] [lim_{x\rightarrow2}\frac{4-\sqrt{18-x}}{x-2}\cdot\frac{4+\sqrt{18-x}}{4+\sqrt{18-x}}=] [lim_{x\rightarrow2}\frac{16-(18-x)}{(x-2)(4+\sqrt{18-x})}=lim_{x\rightarrow2}\frac{x-2}{(x-2)(4+\sqrt{18-x})}=] [lim_{x\rightarrow2}\frac{1}{4+\sqrt{18-x}}=\boxed{\frac{1}{8}}] [lim_{x\rightarrow-3}\frac{\sqrt{x+7}-2}{x+3}=\frac{0}{0}] [lim_{x\rightarrow-3}\frac{\sqrt{x+7}-2}{x+3}\cdot\frac{\sqrt{x+7}+2}{\sqrt{x+7}+2}=] [lim_{x\rightarrow-3}\frac{x+7-4}{x+3(\sqrt{x+7}+2)}=lim_{x\rightarrow-3}\frac{x+3}{(x+3)(\sqrt{x+7}+2)}=] [lim_{x\rightarrow-3}\frac{1}{\sqrt{x+7}+2}=\boxed{\frac{1}{4}}] [lim_{x\rightarrow9}\frac{3-\sqrt{x}}{x-9}=\frac{0}{0}] [lim_{x\rightarrow9}\frac{(3-\sqrt{x})(3+\sqrt{x})}{(x-9)(3+\sqrt{x})}=] [lim_{x\rightarrow9}\frac{9-x}{(x-9)(3+\sqrt{x})}=lim_{x\rightarrow9}-\frac{1(x-9)}{(x-9)(3+\sqrt{x})}=] [lim_{x\rightarrow9}\frac{-1}{3+\sqrt{x}}=\boxed{-\frac{1}{6}}] [lim_{x\rightarrow0}\frac{\sqrt{2x+1}-1}{x}=\frac{0}{0}] [lim_{x\rightarrow0}\frac{\sqrt{2x+1}-1}{x}\cdot\frac{\sqrt{2x+1}+1}{\sqrt{2x+1}+1}=] [lim_{x\rightarrow0}\frac{2x+1-1}{x(\sqrt{2x+1}+1)}=lim_{x\rightarrow0}\frac{2x}{x(\sqrt{2x+1}+1)}=] [lim_{x\rightarrow0}\frac{2}{\sqrt{2x+1}+1}=\boxed{1}] 7. lim_{x→1} (√3 - √(x+2)) / (1-x) lim_{x→1} (√3 - √(x+2)) / (1-x) * (√3 + √(x+2)) / (√3 + √(x+2)) lim_{x→1} (3 - (x+2)) / ((1-x)(√3 + √(x+2))) lim_{x→1} (-x+1) / ((1-x)(√3 + √(x+2))) lim_{x→1} 1 / (√3 + √(x+2)) = √3/6 8. lim_{x→0} (x^(-1) - 1) / (x^(-1) - 1) lim_{x→0} (1/x - 1) / (1/x - 1) lim_{x→0} (1 - x) / x = -1 9. lim_{x→7} (x - 7) / (x^2 - 7x) lim_{x→7} (x - 7) / (x(x - 7)) lim_{x→7} 1 / x = 1/7 10. lim_{x→2} [(x+1)^2 - 1] / (x+2) lim_{x→2} (x^2 + 2x + 1 - 1) / (x+2) lim_{x→2} (x^2 + 2x) / (x+2) lim_{x→2} x(x+2) / (x+2) lim_{x→2} x = 2 11. lim_{x→1} (x^(-4) - 1) / (x-1) lim_{x→1} (1/x^4 - 1) / (x-1) lim_{x→1} (1 - x^4) / (x^4(x-1)) lim_{x→1} (1 + x^2)(1 - x^2) / (x^4(x-1)) lim_{x→1} (1 + x^2)(1 + x)(1 - x) / (x^4(x-1)) lim_{x→1} -(1 + x^2)(1 + x) / x^4 = -4