# Find the limit: lim_{xrightarrowinfty}frac{3x^6-x^4+2x^2-3}{4x^5+2x^3-5}

Find the limit:
$\underset{x\to \mathrm{\infty }}{lim}\frac{3{x}^{6}-{x}^{4}+2{x}^{2}-3}{4{x}^{5}+2{x}^{3}-5}$
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Tuthornt
$\underset{x\to \mathrm{\infty }}{lim}\frac{3{x}^{6}-{x}^{4}+2{x}^{2}-3}{4{x}^{5}+2{x}^{3}-5}$
Multiply and divided by ${x}^{5}$:
$\underset{x\to \mathrm{\infty }}{lim}\frac{3{x}^{6}-{x}^{4}+2{x}^{2}-3}{4{x}^{5}+2{x}^{3}-5}=\underset{x\to \mathrm{\infty }}{lim}\frac{{x}^{5}\frac{3{x}^{6}-{x}^{4}+2{x}^{2}-3}{{x}^{5}}}{{x}^{5}\frac{4{x}^{5}+2{x}^{3}-5}{{x}^{5}}}$
Divide:
$\underset{x\to \mathrm{\infty }}{lim}\frac{{x}^{5}\frac{3{x}^{6}-{x}^{4}+2{x}^{2}-3}{{x}^{5}}}{{x}^{5}\frac{4{x}^{5}+2{x}^{3}-5}{{x}^{5}}}=\underset{x\to \mathrm{\infty }}{lim}\frac{3x-\frac{1}{x}+\frac{2}{{x}^{3}}-\frac{3}{{x}^{5}}}{4+\frac{2}{{x}^{2}}-\frac{5}{{x}^{5}}}$
The limit of the quotient is the quotient of limits:
$\underset{x\to \mathrm{\infty }}{lim}\frac{3x-\frac{1}{x}+\frac{2}{{x}^{3}}-\frac{3}{{x}^{5}}}{4+\frac{2}{{x}^{2}}-\frac{5}{{x}^{5}}}=\frac{\underset{x\to \mathrm{\infty }}{lim}\left(3x-\frac{1}{x}+\frac{2}{{x}^{3}}-\frac{3}{{x}^{5}}\right)}{\underset{x\to \mathrm{\infty }}{lim}\left(4+\frac{2}{{x}^{2}}-\frac{5}{{x}^{5}}\right)}$
The limit of a sum/difference is the sum/difference of limits: $\frac{\underset{x\to \mathrm{\infty }}{lim}\left(3x-\frac{1}{x}+\frac{2}{{x}^{3}}-\frac{3}{{x}^{5}}\right)}{\underset{x\to \mathrm{\infty }}{lim}\left(4+\frac{2}{{x}^{2}}-\frac{5}{{x}^{5}}\right)}=\frac{\left(-\underset{x\to \mathrm{\infty }}{lim}\frac{3}{{x}^{5}}+\underset{x\to \mathrm{\infty }}{lim}\frac{2}{{x}^{3}}-\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}+\underset{x\to \mathrm{\infty }}{lim}3x\right)}{\underset{x\to \mathrm{\infty }}{lim}\left(4+\frac{2}{{x}^{2}}-\frac{5}{{x}^{5}}\right)}$
The limit of a quotient is the quotient of limits:
$\frac{\left(-\underset{x\to \mathrm{\infty }}{lim}\frac{3}{{x}^{5}}+\underset{x\to \mathrm{\infty }}{lim}\frac{2}{{x}^{3}}-\underset{x\to \mathrm{\infty }}{lim}\frac{1}{x}+\underset{x\to \mathrm{\infty }}{lim}3x\right)}{\underset{x\to \mathrm{\infty }}{lim}\left(4+\frac{2}{{x}^{2}}-\frac{5}{{x}^{5}}\right)}=\frac{\left(-\underset{x\to \mathrm{\infty }}{lim}\frac{3}{{x}^{5}}+\underset{x\to \mathrm{\infty }}{lim}\frac{2}{{x}^{3}}-\underset{x\to \mathrm{\infty }}{lim}3x\right)-\frac{\underset{x\to \mathrm{\infty }}{lim}1}{\underset{x\to \mathrm{\infty }}{lim}x}}{\underset{x\to \mathrm{\infty }}{lim}\left(4+\frac{2}{{x}^{2}}-\frac{5}{{x}^{5}}\right)}$
The limit of a constant is equal to the constant: