Question

1. evaluate the limits at infinity.

a) \\( \lim _ { x \
ightarrow \infty } \frac { x ^ { 4 } - 3 x ^ { 3 } + 1 } { x ^ { 3 } - 2 x ^ { 4 } + 2 x } = \\)
b) \\( \lim _ { x \
ightarrow - \infty } \frac { x ^ { 3 } + 7 x - 9 } { x ^ { 2 } - 5 x + 6 } = \\)
c) \\( \lim _ { x \
ightarrow \infty } \frac { ( x ^ { 2 } + 5 x + 1 ) ( x + 2 ) } { x ^ { 4 } - 2 x ^ { 2 } + 2 x } = \\)
d) \\( \lim _ { x \
ightarrow \infty } \frac { 2 x + 3 } { x + \sqrt { 4 x ^ { 2 } + 3 } } = \\)
\\( \lim _ { x \
ightarrow - \infty } \frac { 2 x + 3 } { x + \sqrt { 4 x ^ { 2 } + 3 } } = \\)

Explanation:

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

For \(\lim_{x ightarrow\infty}\frac{x^{4}-3x^{3}+1}{x^{3}-2x^{4}+2x}\), divide numerator and denominator by \(x^{4}\):
\[ \begin{align*} \lim_{x ightarrow\infty}\frac{x^{4}-3x^{3}+1}{x^{3}-2x^{4}+2x}&=\lim_{x ightarrow\infty}\frac{\frac{x^{4}}{x^{4}}-\frac{3x^{3}}{x^{4}}+\frac{1}{x^{4}}}{\frac{x^{3}}{x^{4}}-\frac{2x^{4}}{x^{4}}+\frac{2x}{x^{4}}}\\ &=\lim_{x ightarrow\infty}\frac{1-\frac{3}{x}+\frac{1}{x^{4}}}{\frac{1}{x}-2+\frac{2}{x^{3}}} \end{align*} \]

Step2: Apply limit rules

As \(x ightarrow\infty\), \(\lim_{x ightarrow\infty}\frac{1}{x} = 0\), \(\lim_{x ightarrow\infty}\frac{1}{x^{3}}=0\), \(\lim_{x ightarrow\infty}\frac{1}{x^{4}} = 0\)
\[ \begin{align*} \lim_{x ightarrow\infty}\frac{1-\frac{3}{x}+\frac{1}{x^{4}}}{\frac{1}{x}-2+\frac{2}{x^{3}}}&=\frac{\lim_{x ightarrow\infty}(1)-3\lim_{x ightarrow\infty}(\frac{1}{x})+\lim_{x ightarrow\infty}(\frac{1}{x^{4}})}{\lim_{x ightarrow\infty}(\frac{1}{x})-\lim_{x ightarrow\infty}(2)+2\lim_{x ightarrow\infty}(\frac{1}{x^{3}})}\\ &=\frac{1 - 0+0}{0 - 2+0}=-\frac{1}{2} \end{align*} \]

Step3: For \(\lim_{x

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

Step4: For \(\lim_{x

ightarrow\infty}\frac{(x^{2}+5x + 1)(x + 2)}{x^{4}-2x^{2}+2x}\)
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 and denominator by \(x^{4}\):
\[ \begin{align*} \lim_{x ightarrow\infty}\frac{x^{3}+7x^{2}+11x + 2}{x^{4}-2x^{2}+2x}&=\lim_{x ightarrow\infty}\frac{\frac{x^{3}}{x^{4}}+\frac{7x^{2}}{x^{4}}+\frac{11x}{x^{4}}+\frac{2}{x^{4}}}{\frac{x^{4}}{x^{4}}-\frac{2x^{2}}{x^{4}}+\frac{2x}{x^{4}}}\\ &=\lim_{x ightarrow\infty}\frac{\frac{1}{x}+\frac{7}{x^{2}}+\frac{11}{x^{3}}+\frac{2}{x^{4}}}{1-\frac{2}{x^{2}}+\frac{2}{x^{3}}} \end{align*} \]
As \(x ightarrow\infty\), \(\lim_{x ightarrow\infty}\frac{1}{x}=0\), \(\lim_{x ightarrow\infty}\frac{1}{x^{2}} = 0\), \(\lim_{x ightarrow\infty}\frac{1}{x^{3}}=0\), \(\lim_{x ightarrow\infty}\frac{1}{x^{4}} = 0\)
\[ \begin{align*} \lim_{x ightarrow\infty}\frac{\frac{1}{x}+\frac{7}{x^{2}}+\frac{11}{x^{3}}+\frac{2}{x^{4}}}{1-\frac{2}{x^{2}}+\frac{2}{x^{3}}}&=\frac{0 + 0+0+0}{1-0+0}=0 \end{align*} \]

Step5: For \(\lim_{x

ightarrow\infty}\frac{2x+3}{x+\sqrt{4x^{2}+3}}\)
Divide numerator and denominator by \(x\) (since \(x>0\) as \(x ightarrow\infty\), \(\sqrt{x^{2}}=x\)):
\[ \begin{align*} \lim_{x ightarrow\infty}\frac{2x+3}{x+\sqrt{4x^{2}+3}}&=\lim_{x ightarrow\infty}\frac{2+\frac{3}{x}}{1+\sqrt{4+\frac{3}{x^{2}}}}\\ &=\frac{2 + 0}{1+\sqrt{4+0}}=\frac{2}{1 + 2}=\frac{2}{3} \end{align*} \]
For \(\lim_{x ightarrow-\infty}\frac{2x+3}{x+\sqrt{4x^{2}+3}}\), divide numerator and denominator by \(x\) (since \(x<0\) as \(x ightarrow-\infty\), \(\sqrt{x^{2}}=-x\)):
\[ \begin{align*} \lim_{x ightarrow-\infty}\frac{2x+3}{x+\sqrt{4x^{2}+3}}&=\lim_{x ightarrow-\infty}\frac{2+\frac{3}{x}}{1+\sqrt{4+\frac{3}{x^{2}}}\frac{\sqrt{x^{2}}}{x}}\\ &=\lim_{x ightarrow-\infty}\frac{2+\frac{3}{x}}{1-\sqrt{4+\frac{3}{x^{2}}}}\\ &=\frac{2+0}{1-\sqrt{4+0}}=\frac{2}{1-2}=- 2 \end{align*} \]

Answer:

a) \(-\frac{1}{2}\)
b) \(-\infty\)
c) \(0\)
d) \(\frac{2}{3}\), \(-2\)