# Calculating Limit of Function – A quotient of polynomials in power of a polynomial to infinity – Exercise 6307

### Exercise

Evaluate the following limit:

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

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

### Solution

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

$${(\frac{\infty^2+2\infty-1}{3\infty^2-3\infty-2})}^{\frac{1}{\infty}}$$

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 set of polynomials tending to infinity. In such a situation, we divide the numerator and denominator by the expression with highest power, without its coefficient. We will get:

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

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

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

We plug in infinity again and get

$$={(\frac{1+\frac{2}{\infty}-\frac{1}{\infty^2}}{3-\frac{3}{\infty}-\frac{2}{\infty^2}})}^{\frac{1}{\infty}}=$$

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

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

$$=1$$

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

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