3. evaluate the limits at infinity.\na) \\( \\lim _ { x \\rightarrow \\infty } \\frac { x ^ { 4 } - 3 x ^ {…

Explanation:

Step1: Divide numerator and denominator by highest - power of (x)

For part (a): Divide numerator (x^{4}-3x^{3}+1) and denominator (x^{3}-2x^{4}+2x) by (x^{4}). [ \begin{align*} \lim_{x\rightarrow\infty}\frac{x^{4}-3x^{3}+1}{x^{3}-2x^{4}+2x}&=\lim_{x\rightarrow\infty}\frac{\frac{x^{4}}{x^{4}}-\frac{3x^{3}}{x^{4}}+\frac{1}{x^{4}}}{\frac{x^{3}}{x^{4}}-\frac{2x^{4}}{x^{4}}+\frac{2x}{x^{4}}}\ &=\lim_{x\rightarrow\infty}\frac{1 - \frac{3}{x}+\frac{1}{x^{4}}}{\frac{1}{x}-2+\frac{2}{x^{3}}} \end{align*} ]

Step2: Use (\lim_{x\rightarrow\infty}\frac{1}{x^{n}} = 0) for (n>0)

As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{1}{x}=0), (\lim_{x\rightarrow\infty}\frac{1}{x^{3}} = 0), (\lim_{x\rightarrow\infty}\frac{1}{x^{4}}=0) [ \lim_{x\rightarrow\infty}\frac{1 - \frac{3}{x}+\frac{1}{x^{4}}}{\frac{1}{x}-2+\frac{2}{x^{3}}}=\frac{1-0 + 0}{0 - 2+0}=-\frac{1}{2} ]

Step3: For part (b)

Divide numerator (x^{3}+7x - 9) and denominator (x^{2}-5x + 6) by (x^{2}) [ \begin{align*} \lim_{x\rightarrow-\infty}\frac{x^{3}+7x - 9}{x^{2}-5x + 6}&=\lim_{x\rightarrow-\infty}\frac{\frac{x^{3}}{x^{2}}+\frac{7x}{x^{2}}-\frac{9}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{5x}{x^{2}}+\frac{6}{x^{2}}}\ &=\lim_{x\rightarrow-\infty}\frac{x+\frac{7}{x}-\frac{9}{x^{2}}}{1-\frac{5}{x}+\frac{6}{x^{2}}} \end{align*} ] As (x\rightarrow-\infty), (\lim_{x\rightarrow-\infty}\frac{1}{x}=0), (\lim_{x\rightarrow-\infty}\frac{1}{x^{2}} = 0) [ \lim_{x\rightarrow-\infty}\frac{x+\frac{7}{x}-\frac{9}{x^{2}}}{1-\frac{5}{x}+\frac{6}{x^{2}}}=-\infty ]

Step4: For part (c)

First, expand the numerator ((x^{2}+5x + 1)(x + 2)=x^{3}+2x^{2}+5x^{2}+10x+x + 2=x^{3}+7x^{2}+11x + 2) Divide numerator (x^{3}+7x^{2}+11x + 2) and denominator (x^{4}-2x^{2}+2x) by (x^{4}) [ \begin{align*} \lim_{x\rightarrow\infty}\frac{(x^{2}+5x + 1)(x + 2)}{x^{4}-2x^{2}+2x}&=\lim_{x\rightarrow\infty}\frac{\frac{x^{3}}{x^{4}}+\frac{7x^{2}}{x^{4}}+\frac{11x}{x^{4}}+\frac{2}{x^{4}}}{\frac{x^{4}}{x^{4}}-\frac{2x^{2}}{x^{4}}+\frac{2x}{x^{4}}}\ &=\lim_{x\rightarrow\infty}\frac{\frac{1}{x}+\frac{7}{x^{2}}+\frac{11}{x^{3}}+\frac{2}{x^{4}}}{1-\frac{2}{x^{2}}+\frac{2}{x^{3}}} \end{align*} ] Using (\lim_{x\rightarrow\infty}\frac{1}{x^{n}}=0) for (n > 0), we get (\lim_{x\rightarrow\infty}\frac{\frac{1}{x}+\frac{7}{x^{2}}+\frac{11}{x^{3}}+\frac{2}{x^{4}}}{1-\frac{2}{x^{2}}+\frac{2}{x^{3}}}=0)

Step5: For part (d) (when (x\rightarrow\infty))

Divide numerator and denominator by (x) ((x>0), so (\sqrt{x^{2}}=x)) [ \begin{align*} \lim_{x\rightarrow\infty}\frac{2x + 3}{x+\sqrt{4x^{2}+3}}&=\lim_{x\rightarrow\infty}\frac{2+\frac{3}{x}}{1+\sqrt{4+\frac{3}{x^{2}}}}\ \end{align*} ] Using (\lim_{x\rightarrow\infty}\frac{1}{x}=0) and (\lim_{x\rightarrow\infty}\frac{1}{x^{2}}=0) [ \lim_{x\rightarrow\infty}\frac{2+\frac{3}{x}}{1+\sqrt{4+\frac{3}{x^{2}}}}=\frac{2 + 0}{1+\sqrt{4+0}}=\frac{2}{3} ]

Step6: For part (d) (when (x\rightarrow-\infty))

Divide numerator and denominator by (x) ((x<0), so (\sqrt{x^{2}}=-x)) [ \begin{align*} \lim_{x\rightarrow-\infty}\frac{2x + 3}{x+\sqrt{4x^{2}+3}}&=\lim_{x\rightarrow-\infty}\frac{2+\frac{3}{x}}{1-\sqrt{4+\frac{3}{x^{2}}}}\ \end{align*} ] Using (\lim_{x\rightarrow-\infty}\frac{1}{x}=0) and (\lim_{x\rightarrow-\infty}\frac{1}{x^{2}}=0) [ \lim_{x\rightarrow-\infty}\frac{2+\frac{3}{x}}{1-\sqrt{4+\frac{3}{x^{2}}}}=\frac{2+0}{1 - 2}=- 2 ]

Answer:

a) (-\frac{1}{2}) b) (-\infty) c) (0) d) When (x\rightarrow\infty), (\frac{2}{3}); when (x\rightarrow-\infty), (-2)