Following the hint, we have
lim x → − ∞ x 4 + x 3 + 2 − x 2 5 x + 3 = lim x → − ∞ x 4 + x 3 + 2 − x 2 5 x + 3 ⋅ x 4 + x 3 + 2 + x 2 x 4 + x 3 + 2 + x 2 = lim x → − ∞ x 4 + x 3 + 2 − x 4 ( 5 x + 3 ) ⋅ ( x 4 + x 3 + 2 + x 2 ) = lim x → − ∞ x 3 + 2 ( 5 x + 3 ) ( x 4 + x 3 + 2 + x 2 ) . {\displaystyle {\begin{aligned}&\lim _{x\to -\infty }{\frac {{\sqrt {x^{4}+x^{3}+2}}-x^{2}}{5x+3}}\\&=\lim _{x\to -\infty }{\frac {{\sqrt {x^{4}+x^{3}+2}}-x^{2}}{5x+3}}\cdot {\frac {{\sqrt {x^{4}+x^{3}+2}}+x^{2}}{{\sqrt {x^{4}+x^{3}+2}}+x^{2}}}\\&=\lim _{x\to -\infty }{\frac {x^{4}+x^{3}+2-x^{4}}{(5x+3)\cdot ({\sqrt {x^{4}+x^{3}+2}}+x^{2})}}\\&=\lim _{x\to -\infty }{\frac {x^{3}+2}{(5x+3)({\sqrt {x^{4}+x^{3}+2}}+x^{2})}}.\end{aligned}}}
Now if we divide both the denominator and the numerator by x 3 {\displaystyle x^{3}} , we get
lim x → − ∞ x 4 + x 3 + 2 − x 2 5 x + 3 = lim x → − ∞ x 3 + 2 x 3 5 x + 3 x ⋅ x 4 + x 3 + 2 + x 2 x 2 = lim x → − ∞ 1 + 2 x 3 ( 5 + 3 x ) ⋅ ( x 4 + x 3 + 2 x 4 + x 2 x 2 ) = lim x → − ∞ 1 + 2 x 3 ( 5 + 3 x ) ⋅ ( 1 + 1 x + 2 x 4 + 1 ) = 1 5 ⋅ ( 1 + 1 ) = 1 10 , {\displaystyle {\begin{aligned}&\lim _{x\to -\infty }{\frac {{\sqrt {x^{4}+x^{3}+2}}-x^{2}}{5x+3}}\\&=\lim _{x\to -\infty }{\frac {\frac {x^{3}+2}{x^{3}}}{{\frac {5x+3}{x}}\cdot {\frac {{\sqrt {x^{4}+x^{3}+2}}+x^{2}}{x^{2}}}}}\\&=\lim _{x\to -\infty }{\frac {1+{\frac {2}{x^{3}}}}{\left(5+{\frac {3}{x}}\right)\cdot \left({\sqrt {\frac {x^{4}+x^{3}+2}{x^{4}}}}+{\frac {x^{2}}{x^{2}}}\right)}}\\&=\lim _{x\to -\infty }{\frac {1+{\frac {2}{x^{3}}}}{\left(5+{\frac {3}{x}}\right)\cdot \left({\sqrt {1+{\frac {1}{x}}+{\frac {2}{x^{4}}}}}+1\right)}}\\&={\frac {1}{5\cdot (1+1)}}\\&={\frac {1}{10}},\end{aligned}}}
so lim x → − ∞ x 4 + x 3 + 2 − x 2 5 x + 3 = 1 10 {\displaystyle \color {blue}\lim _{x\to -\infty }{\frac {{\sqrt {x^{4}+x^{3}+2}}-x^{2}}{5x+3}}={\frac {1}{10}}} .
, we get
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