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

Exercise

Evaluate the following limit:

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

Final Answer

Show final answer

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

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!