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I have to calculate

$$\lim_{x \to +\infty} \left(\dfrac{2x-5}{2x-2}\right)^{4x^2}$$

After having arranged it as

$$\lim_{x \to +\infty} \left(1 - \dfrac{3}{2x-2}\right)^{4x^2}$$

I used $2x-2 = t$ thence

$$\lim_{t\to +\infty} \left(1 - \dfrac{3}{t}\right)^{(t+2)^2}$$

Is now legit to solve it like this?

$$\lim_{t\to +\infty} \left(\underbrace{\left(1-\dfrac{3}{t}\right)^t}_{e^{-3}}\right)^t\cdot \lim_{t\to +\infty}\underbrace{\left(1-\dfrac{3}{t}\right)^{4t}}_{e^{-12}}\cdot \lim_{t\to +\infty}\underbrace{\left(1-\dfrac{3}{t}\right)^4}_{1}$$

Hence the limit is zero.

  • Is this legit? Why not, if? I mean the separation into three products of limits.
Gerr
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  • The justification for the first limit is missing. Otherwise it would be okay – Brian Moehring Oct 23 '24 at 15:38
  • Yes if all the separated limits exist. – Bowei Tang Oct 23 '24 at 15:38
  • @BrianMoehring Yes I went a bit fast. It's zero because it's $e^{-3t}$ which goes to zero as $x \to +\infty$. That was my principal concern, actually, the possibility to solve a limit "inside a limit" – Gerr Oct 23 '24 at 16:11
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    @J.N. The problem is that it's never actually $e^{-3t}$. Rather, it's $(\alpha_t)^t$ where $\alpha_t \to e^{-3}$. This means, for instance, that there's some $t_0$ such that $t \geq t_0$ implies $0 \leq \alpha_t \leq e^{-2}$, and therefore $0 \leq (\alpha_t)^t \leq e^{-2t}$. Now, the fact that $e^{-2t} \to 0$ allows us to imply the original limit also tends to $0$. – Brian Moehring Oct 24 '24 at 14:26
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    A final note: The above argument is simple enough that I'd never require a professional to make it. It's also close enough to a mistake students regularly make that, if you're still developing your intuition, you should go through these steps to explain to yourself why you can do this. – Brian Moehring Oct 24 '24 at 14:31

3 Answers3

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Yes indeed algebrically we have

$$\left(1 - \dfrac{3}{t}\right)^{(t+2)^2}=\left(1 - \dfrac{3}{t}\right)^{t^2}\left(1 - \dfrac{3}{t}\right)^{4t}\left(1 - \dfrac{3}{t}\right)^{4} \to 0\cdot e^{-12}\cdot 1=0$$

and product rule for limits holds.

As an alternative we can use that as $x \to 0$, $\log(1+x)\le x$ then

$$\log\left[\left(1 - \dfrac{3}{t}\right)^{(t+2)^2}\right]=(t+2)^2\log \left(1 - \dfrac{3}{t}\right)\le -\frac{3(t+2)^2}t \to -\infty$$

user
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3

Note that $0\le 1-\frac 3t\le 1$

Therefore $0\le(1-\frac 3t)^\alpha\le 1$ for any power $\alpha>0$ and you can just ignore the $4t+4$ exponent with this rough inequality.

$$0\le(1-\tfrac 3t)^{(t+2)^2}\le(1-\tfrac 3t)^{t^2}\to 0$$

zwim
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The three limits exist so yes. In general, $\lim_{t\to\infty} f(t)\cdot g(t)=\lim_{t\to\infty} f(t) \cdot \lim_{t\to\infty} g(t)$ if both limits exist.

user1360376
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