# Solve the following limit. \lim_{x\to\infty}(\frac{3x^3-4x+2}{7x^3+5})

Solve the following limit.
$$\displaystyle\lim_{{{x}\to\infty}}{\left({\frac{{{3}{x}^{{3}}-{4}{x}+{2}}}{{{7}{x}^{{3}}+{5}}}}\right)}$$

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izboknil3
Divide numerator and denominator of the expression with the highest power of x in the denominator.
Thus we can write the limit expression as,
$$\displaystyle\lim_{{{x}\to\infty}}{\left({\frac{{{3}{x}^{{3}}-{4}{x}+{2}}}{{{7}{x}^{{3}}+{5}}}}\right)}=\lim_{{{x}\to\infty}}{\left({\frac{{{3}{\frac{{{x}^{{3}}}}{{{x}^{{3}}}}}-{4}{\frac{{{x}}}{{{x}^{{3}}}}}+{\frac{{{2}}}{{{x}}}}}}{{{7}{\frac{{{x}^{{3}}}}{{{x}^{{3}}}}}+{\frac{{{5}}}{{{x}^{{3}}}}}}}}\right)}$$
$$\displaystyle=\lim_{{{x}\to\infty}}{\left({\frac{{{3}-{\frac{{{4}}}{{{x}^{{2}}}}}+{\frac{{{2}}}{{{x}^{{3}}}}}}}{{{7}+{\frac{{{5}}}{{{x}^{{3}}}}}}}}\right)}$$
Now simplify the limit expression using limit properties and apply the limit value to the variable x.
The terms, $$\displaystyle\lim_{{{x}\to\infty}}{\frac{{{4}}}{{{x}^{{2}}}}},\lim_{{{x}\to\infty}}{\frac{{{2}}}{{{x}^{{3}}}}},\lim_{{{x}\to\infty}}{\frac{{{5}}}{{{x}^{{3}}}}}$$ approaches 0 as x approaches infinity. In other words,
$$\displaystyle\lim_{{{x}\to\infty}}{\frac{{{4}}}{{{x}^{{2}}}}}={\frac{{{4}}}{{\infty}}}$$
$$\displaystyle={\frac{{{4}}}{{{\frac{{{1}}}{{{0}}}}}}}$$
$$\displaystyle={4}\times{\frac{{{0}}}{{{1}}}}$$
$$\displaystyle={0}$$
Similarly we get all the terms $$\displaystyle\lim_{{{x}\to\infty}}{\frac{{{4}}}{{{x}^{{2}}}}},\lim_{{{x}\to\infty}}{\frac{{{2}}}{{{x}^{{3}}}}},\lim_{{{x}\to\infty}}{\frac{{{5}}}{{{x}^{{3}}}}}$$ equal to 0.
Therefore we get,
$$\displaystyle\lim_{{{x}\to\infty}}{\left({\frac{{{3}-{\frac{{{4}}}{{{x}^{{2}}}}}+{\frac{{{2}}}{{{x}^{{3}}}}}}}{{{7}+{\frac{{{5}}}{{{x}^{{3}}}}}}}}\right)}={\frac{{\lim_{{{x}\to\infty}}{\left({3}-{\frac{{{4}}}{{{x}^{{2}}}}}+{\frac{{{2}}}{{{x}^{{3}}}}}\right)}}}{{\lim_{{{x}\to\infty}}{\left({7}+{\frac{{{5}}}{{{x}^{{3}}}}}\right)}}}}$$
$$\displaystyle={\frac{{{3}-\lim_{{{x}\to\infty}}{\frac{{{4}}}{{{x}^{{2}}}}}+\lim_{{{x}\to\infty}}{\frac{{{2}}}{{{x}^{{3}}}}}}}{{{7}+\lim_{{{x}\to\infty}}{\frac{{{5}}}{{{x}^{{3}}}}}}}}$$
$$\displaystyle={\frac{{{3}-{0}+{0}}}{{{7}+{0}}}}={\frac{{{3}}}{{{7}}}}$$
Hence the limit of the given expression os $$\displaystyle{\frac{{{3}}}{{{7}}}}$$