Facing difficulty finding limit
$$\lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}$$
For starters I have trouble simplifying it
Which method would help in finding this limit?
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1$\begingroup$ This is already solved, but another useful hint: $$\frac{x}{x-1} = \frac{x-1+1}{x-1} = 1+\frac{1}{x-1}.$$ $\endgroup$– JavaManCommented Dec 11, 2011 at 5:11
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6 Answers
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If you know that $$\lim_{x\to\infty}\left(1 + \frac{a}{x}\right)^x = e^{a},$$ so that $$\lim_{x\to\infty}\left(1 - \frac{1}{x}\right)^x = e^{-1},$$ then you can try to rewrite your limit into something involving this limit.
So try rewriting it; perhaps as a product, $$\begin{align*} \left(\frac{x}{x-1}\right)^{2x+1} &= \left(\left(\frac{x}{x-1}\right)^x\right)^2\left(\frac{x}{x-1}\right)\\ &= \left(\frac{1}{\left(\frac{x-1}{x}\right)^x}\right)^2\left(\frac{x}{x-1}\right)\\ &= \left(\frac{1}{\left(1 - \frac{1}{x}\right)^x}\right)^2\left(\frac{x}{x-1}\right). \end{align*}$$ Then use limit laws to compute it.
answered Dec 11, 2011 at 0:54
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$\begingroup$ I love this answer! Very elegant! $\endgroup$ Commented Dec 11, 2011 at 1:08
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1$\begingroup$ An initial substitution of y=x-1 makes it a lot easier. $\endgroup$– tzsCommented Dec 11, 2011 at 4:45
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1$\begingroup$ @tzs maybe you mean $y = x+1$. Check out my answer. $\endgroup$ Commented Dec 11, 2011 at 5:00
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First, try finding the limit of its logarithm. If you write $(2x+1)\cdot(\text{something}) = \frac{\text{something}}{1/(2x+1)}$, then L'Hopital's rule should do it. Then take the antilogarithm of that and you've got it.
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This is going to be very similar to what Arturo suggested but it has the benefit of arriving at the answer quicker. Using a substitution $x \mapsto y+1$ we can write the function as
$$\left(\frac{x}{x-1}\right)^{2x+1} = \left(1+\frac{1}{y}\right)^{2y+3} =\left(\left(1+\frac{1}{y}\right)^y\right)^2\left(1+\frac{1}{y}\right)^{3}$$
Finding the limit of this one should be easy
answered Dec 11, 2011 at 4:43
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$$ \begin{eqnarray} \lim \limits_{x\to \infty}\left(\frac{x}{x-1}\right)^{2x+1}=\lim \limits_{x\to \infty}\left(\frac{x-1+1}{x-1}\right)^{2x+1} =\lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{2x+1}\\= \lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{(x-1)\cdot\frac{2x+1}{x-1}} =\lim \limits_{x\to \infty}\left(1+\frac{1}{x-1}\right)^{(x-1)\cdot\frac{2x+1}{x-1}}=e^{\lim \limits_{x\to \infty}\frac{2x+1}{x-1}}=e^2 \end{eqnarray} $$
answered Jul 5, 2012 at 7:52
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$\lim_{x\to\infty}(\frac{x}{x-1})^{2x+1}=\lim_{x\to\infty}(\frac{x-1+1}{x-1})^{2x+1}=\lim_{x\to\infty}(1+\frac{1}{x-1})^{2x+1}=e^{\lim_{x\to\infty}\frac{2x+1}{x-1}}=e^2$.
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Limits of the type f(x)^g(x) where f(x) tends to 1 and g(x) to infinity i.e 1^infinity
can be done as e^((f(x)-1).g(x))
if you do that you get e^2
