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


Evaluate the following limit:

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

Final Answer

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


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


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





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! 

Share with Friends

Leave a Reply