# Calculating Limit of Function – A quotient of functions with square roots to infinity – Exercise 6570

### Exercise

Evaluate the following limit:

$$\lim _ { x \rightarrow \infty} \frac{\sqrt[6]{x^3-\sqrt{x}}}{\sqrt{\sqrt[6]{x}+3\sqrt[3]{x}+4x}+\sqrt[8]{x+x^4}}$$

$$\lim _ { x \rightarrow \infty} \frac{\sqrt[6]{x^3-\sqrt{x}}}{\sqrt{\sqrt[6]{x}+3\sqrt[3]{x}+4x}+\sqrt[8]{x+x^4}}=\frac{1}{3}$$

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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