Exercise
Evaluate the following limit:
\lim _ { x \rightarrow \infty} \frac{\sqrt[6]{x^3-\sqrt{x}}}{\sqrt{\sqrt[6]{x}+3\sqrt[3]{x}+4x}+\sqrt[8]{x+x^4}}
Final Answer
Solution
First, we try to plug in x = \infty and get
\frac{\sqrt[6]{\infty^3-\sqrt{\infty}}}{\sqrt{\sqrt[6]{\infty}+3\sqrt[3]{\infty}+4\infty}+\sqrt[8]{\infty+\infty^4}}=\frac{\infty-\infty}{\infty}
We got the phrase \infty-\infty (=infinity minus infinity). This is an indeterminate form, therefore we have to get out of this situation.
We divide numerator and denominator by the leading factor (= the expression with the highest power). In this case, it will be the expression:
\sqrt{x}
Division gives us the following:
\lim _ { x \rightarrow \infty} \frac{\sqrt[6]{x^3-\sqrt{x}}}{\sqrt{\sqrt[6]{x}+3\sqrt[3]{x}+4x}+\sqrt[8]{x+x^4}}=
=\lim _ { x \rightarrow \infty} \frac{\frac{\sqrt[6]{x^3-\sqrt{x}}}{\sqrt{x}}}{\frac{\sqrt{\sqrt[6]{x}+3\sqrt[3]{x}+4x}+\sqrt[8]{x+x^4}}{\sqrt{x}}}=
=\lim _ { x \rightarrow \infty} \frac{\sqrt[6]{\frac{x^3-\sqrt{x}}{x^3}}}{\sqrt{\frac{\sqrt[6]{x}+3\sqrt[3]{x}+4x}{x}}+\sqrt[8]{\frac{x+x^4}{x^4}}}=
=\lim _ { x \rightarrow \infty} \frac{\sqrt[6]{1-\frac{1}{x^{\frac{5}{2}}}}}{\sqrt{\frac{1}{x^{\frac{5}{6}}}+\frac{3}{x^{\frac{2}{3}}}+4}+\sqrt[8]{\frac{1}{x^3}+1}}=
We plug in infinity again and get
=\frac{\sqrt[6]{1-0}}{\sqrt{0+0+4}+\sqrt[8]{0+1}}=
=\frac{1}{2+1}=
=\frac{1}{3}
Note: Infinity in the power of any positive number equals to infinity. Also, an infinite number divides by infinity is defined and equals to zero. For the full list press here
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