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

Evaluate the limit
$\underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2}{3{x}^{4}+5x}$
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Given:
$\underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2}{3{x}^{4}+5x}$
To find:
The limit. $\underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2}{3{x}^{4}+5x}$
Solve:
$\underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2}{3{x}^{4}+5x}$
On multiplying and dividing by ${x}^{4}$
$\underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2}{3{x}^{4}+5x}=\underset{x\to \mathrm{\infty }}{lim}\frac{{x}^{4}\left(\frac{4}{x}-\frac{2}{{x}^{4}}\right)}{{x}^{4}\left(3-\frac{5}{{x}^{3}}\right)}$
On cancelling the common term,
$\underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2}{3{x}^{4}+5x}=\underset{x\to \mathrm{\infty }}{lim}\frac{\left(\frac{4}{x}-\frac{2}{{x}^{4}}\right)}{\left(3-\frac{5}{{x}^{3}}\right)}$
Substitute the variable with value,
$\underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2}{3{x}^{4}+5x}=\underset{x\to \mathrm{\infty }}{lim}\frac{\left(4\left(0\right)-2\left(0\right)\right)}{\left(3-5\left(0\right)\right)}$
$\underset{x\to \mathrm{\infty }}{lim}\frac{4{x}^{3}-2}{3{x}^{4}+5x}=0$