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Find the following limits if they exist:

$$\lim _{x \rightarrow \infty} \frac{4 x^{2}+x+1}{7 x^{2}+3 x-2}$$

$$\lim _{x \rightarrow \infty} \frac{x^{5}+x+3}{x^{3}+x-2}$$

$$\lim _{x \rightarrow \infty} \frac{x^{3}+x+3}{x^{5}+x-2}$$

$$\lim _{x \rightarrow \infty} \frac{4 x^{2}+x+1}{7 x^{2}+3 x-2}$$

$$=\frac{4 x^2 } {7 x^{2}}=\frac{4}{7}$$

$$\lim _{x \rightarrow \infty} \frac{x^{5}+x+3}{x^{3}+x-2}$$


$$\lim _{x \rightarrow \infty} \frac{x^{3}+x+3}{x^{5}+x-2}$$


$$=\frac {1}{x^2}=\frac{1}{\infty}=0$$

Find the following limits if they exist:

$$\lim _{x \rightarrow \infty} \frac{\sqrt{2 x^{2}+1}}{x+1}$$

$$\lim _{x \rightarrow-\infty} \frac{2 x-1}{\sqrt{1+9 x^{2}}}$$

$$\lim _{x \rightarrow \infty} \frac{\sqrt{2 x^{2}+1}}{x+1}=\lim _{x \rightarrow \infty} \frac{\sqrt{x^{2}\left(2+1 / x^{2}\right)}}{x+1}$$

$$=\lim _{x \rightarrow \infty}\frac {|x| \sqrt{2+1 / x^{2}}}{x+1}$$

when $$x \rightarrow \infty \quad ; \quad|x|=x$$

$$\lim _{x \rightarrow \infty} \frac{x \sqrt{2+1 / x^{2}}}{x(1+1 / x)}=\lim _{x \rightarrow \infty} \frac{\sqrt{2+1 / x^{2}}}{1+1 / x}$$

$$=\lim \frac{\sqrt{2+1 / \infty^{2}}}{1+1 / \infty}=\frac{\sqrt{2}}{1}=\sqrt{2}$$

$$\lim _{x \rightarrow-\infty} \frac{2 x-1}{\sqrt{1+9 x^{2}}}=\frac{2(-\infty)-1}{\sqrt{1+9(-\infty)^{2}}}=\frac{-\infty}{\infty}$$

$$\lim _{x \rightarrow-\infty} \frac{2 x-1}{\sqrt{x^{2}\left(1 / x^{2}+9\right)}} =\lim _{x \rightarrow-\infty} \frac{x(2-1 / x)}{ |x|\sqrt{1 / x^{2}+9}}$$

When $$x \rightarrow-\infty \: ; \: |x|=-x  \quad ,\: \frac{1}{\infty}=0$$

$$\lim _{x \rightarrow-\infty} \frac{x(2-1 / x)}{-x \sqrt{1 / x^{2}+9}}$$

$$\lim _{x \rightarrow-\infty} \frac{2-1 / x}{-\sqrt{1 / x^{2}+9}}=\frac{2}{-\sqrt{9}}=\frac{2}{-3}$$

Find the following limit $$\lim _{x \rightarrow-\infty}\left(x+\sqrt{x^{2}+3 x+1}\right)$$

$$\lim _{x \rightarrow-\infty}\left(x+\sqrt{x^{2}+3 x+1}\right)=-\infty+\sqrt{(-\infty)^{2}+3(-\infty)+1}$$


$$\lim _{x \rightarrow-\infty}\left(x+\sqrt{x^{2}+3 x+1}\right) \cdot \frac{x-\sqrt{x^{2}+3 x+1}}{x-\sqrt{x^{2}+3 x+1}}$$

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

When $$x \rightarrow-\infty \: , \:|x|=-x$$
$$\lim _{x \rightarrow-\infty} \frac{-3 x-1}{x-|x| \sqrt{3 / x+1 / x^{2}+1}}=$$

$$\lim _{x \rightarrow \infty} \frac{-3 x-1}{x+x \sqrt{3 / x+1 / x^{2}+1}}$$

$$=\lim _{x \rightarrow-\infty} \frac{x(-3-1 / x)}{x\left(1+\sqrt{1+3 / x+1 / x^{2}}\right)}$$

$$=\lim _{x \rightarrow-\infty} \frac{-3-\frac {1}{x}}{1+\sqrt{1+3 / x+1 / x^{2}}}=\frac{-3}{1+\sqrt{1}}=\frac{-3}{2}$$

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