# Calculating Limit of Function – A quotient of polynomials in the power of a polynomial to infinity – Exercise 6559

### Exercise

Evaluate the following limit:

$$\lim _ { x \rightarrow \infty} {(\frac{3x^2+8x-6}{x^2-5x+2})}^{-x}$$

$$\lim _ { x \rightarrow \infty} {(\frac{3x^2+8x-6}{x^2-5x+2})}^{-x}=0$$

### Solution

First, we try to plug in $$x = \infty$$ and get

$${(\frac{3\infty^2+8\infty-6}{\infty^2-5\infty+2})}^{-\infty}$$

In the base we got the phrase $$\frac{\infty}{\infty}$$ (=infinity divides by infinity). This is an indeterminate form, therefore we have to get out of this situation.

We have a quotient of polynomials tending to infinity. In such a case, we divide the numerator and denominator by the expression with the highest power, without its coefficient. In this case, we get

$$\lim _ { x \rightarrow \infty} {(\frac{3x^2+8x-6}{x^2-5x+2})}^{-x}=$$

$$=\lim _ { x \rightarrow \infty} {(\frac{\frac{3x^2+8x-6}{x^2}}{\frac{x^2-5x+2}{x^2}})}^{-x}=$$

$$=\lim _ { x \rightarrow \infty} {(\frac{3+\frac{8}{x}-\frac{6}{x^2}}{1-\frac{-5}{x}+\frac{2}{x^2}})}^{-x}=$$

We will plug in infinity again and get

$$={(\frac{3+0-0}{1-0+0})}^{-\infty}=$$

$$=3^{-\infty}=$$

$$=\frac{1}{3^{\infty}}=$$

$$=\frac{1}{\infty}=$$

$$=0$$

Note: Any positive finite number divides by infinity is defined and equals to zero. For the full list press here

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