Question

Evaluate :

$\displaystyle\lim_{x \to \infty}\: {(\frac{x^2+2x-1}{x^2-x-4})}^{(2x+1)}$

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Answer 1

Answer:

Step-by-step explanation:

Answer 2

ANSWER

$e^{6}$

Step By Step Explanation

$\displaystyle\lim_{x \to \infty}\: {(\dfrac{x^2+2x-1}{x^2-x-4}) } ^{(2x+1)}$

$(\rightarrow 1) ^{\rightarrow\infty}$ form.

$\implies e^{ \displaystyle \lim_{x \to \infty}\: (\dfrac{x^{2} + 2x -1}{x^{2} - 4 - x } -1)(2x + 1) }$

$\implies e^{ \displaystyle \lim_{x \to \infty}\: (\dfrac{3x + 3}{x^{2} - 4 - x} -1)(2x + 1) }$

Also we know that >>

$\displaystyle\lim_{x \to \infty}\:(\dfrac{1}{x}) = 0$

Now we get ->

$\implies e^{\displaystyle \lim_{x \to \infty}\: \dfrac{ (x^{2}) (3-\dfrac{3}{x} )(2+ \dfrac{1}{x}) } { (x^{2})(1 - \dfrac{1}{x} - \dfrac{4} {x^{2} })}}$

$\implies e^{\frac{(3-0)(2+0) }{1-0-0} }$

$\implies e^{6}$

Hence ,the evaluation of $\displaystyle\lim_{x \to \infty}\: {(\frac{x^2+2x-1}{x^2-x-4})}^ {(2x+1)}$ is $e^{6}$.