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lim_(x rarr oo)(x^(2))/(x^(3)-5)

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

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Evaluate definite integrals using the chain rule

Full solution

Q. i)

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

Divide by $x^3$ : To find the limit of the function as $x$ approaches infinity, we can divide the numerator and the denominator by the highest power of $x$ in the denominator, which is $x^3$ .
Simplify expression: Divide both the numerator and the denominator by $x^3$ : $\lim_{x \to \infty} \left(\frac{x^2}{x^3}\right) / \left(\frac{x^3}{x^3} - \frac{5}{x^3}\right)$
Approach infinity: Simplify the expression: $\lim_{x \to \infty} \frac{1/x}{1 - 5/x^3}$
Final limit is $0$ : As $x$ approaches infinity, $\frac{1}{x}$ and $\frac{5}{x^3}$ both approach $0$ : $\lim_{x \to \infty} \frac{\frac{1}{x}}{1 - \frac{5}{x^3}} = \frac{0}{1 - 0}$
Final limit is $0$ : As $x$ approaches infinity, $\frac{1}{x}$ and $\frac{5}{x^3}$ both approach $0$ : $\lim_{x \to \infty} \left(\frac{1}{x}\right) / \left(1 - \frac{5}{x^3}\right) = \frac{0}{1 - 0}$ The limit of the function as $x$ approaches infinity is $0$ : $\lim_{x \to \infty} \left(\frac{x^2}{x^3 - 5}\right) = 0$

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