Question: "Facing difficulty finding limitlimx→∞(xx−1)2x+1"

Destinee Hensley

Answered question

2022-04-04

Facing difficulty finding limit

lim_{x \to \infty} {(\frac{x}{x - 1})}^{2 x + 1}

membatas0v2v

Beginner2022-04-05Added 19 answers

If you know that

lim_{x \to \infty} {(1 + \frac{a}{x})}^{x} = e^{a}

so that

lim_{x \to \infty} {(1 - \frac{1}{x})}^{x} = e^{- 1}

then you can try to rewrite your limit into something involving this limit.
So try rewriting it; perhaps as a product,

{(\frac{x}{x - 1})}^{2 x + 1} = {({(\frac{x}{x - 1})}^{x})}^{2} (\frac{x}{x - 1})

= {(\frac{1}{{(\frac{x - 1}{x})}^{x}})}^{2} (\frac{x}{x - 1})

= {(\frac{1}{{(1 - \frac{1}{x})}^{x}})}^{2} (\frac{x}{x - 1})

clarkchica44klt

Beginner2022-04-06Added 17 answers

lim_{x \to \infty} {(\frac{x}{x - 1})}^{2 x + 1} = lim_{x \to \infty} {(\frac{x - 1 + 1}{x - 1})}^{2 x + 1}

= lim_{x \to \infty} {(1 + \frac{1}{x - 1})}^{2 x + 1} = lim_{x \to \infty} {(1 + \frac{1}{x - 1})}^{(x - 1) \cdot \frac{2 x + 1}{x - 1}}

= lim_{x \to \infty} {(1 + \frac{1}{x - 1})}^{(x - 1) \cdot \frac{2 x + 1}{x - 1}}

= e^{lim_{x \to \infty} \frac{2 x + 1}{x - 1}} = e^{2}

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