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

Exercise

Evaluate the following limit:

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

Final Answer

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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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