Why is it wrong to solve the limit this way?

Question

Consider the following limit： \begin{align*} &\lim_{x\to +\infty}\frac{e^x}{\left(1+\frac1x\right)^{x^2}}\\ &=e^{\lim_{x\to +\infty}(x-x^2\ln\left(1+\frac1x\right))}\\ &=e^{\lim_{x\to +\infty}x^2(\frac1x-\ln\left(1+\frac1x\right))}\\ &=e^{\lim_{x\to 0^+ }\frac1{x^2}(x-\ln(1+x))}\\ &=e^{\lim_{x\to 0^+ }\frac{\frac12x^2}{x^2}}\\ &=e^{\frac12} \end{align*} The above solution is correct and another solution is following: \begin{align*} &\lim_{x\to +\infty}\frac{e^x}{(1+\frac1x)^{x^2}}\\ &=\lim_{x\to +\infty}\frac{e^x}{[(1+\frac1x)^{x}]^x}\\ &=\lim_{x\to +\infty}\frac{e^x}{e^x}\\ &=1 \end{align*} Why is that wrong? Is that because we didn't take the limit of both the numerator and the denominator?

Answer 1

It is not allowed, in general, take the limit for a single part of the entire expression.

Notably, in this case you have made this mistake twice in two different ways using that

$$\lim_{x\to +\infty}\frac{f(x)}{g(x)} =\lim_{x\to +\infty}\frac{f(x)}{\lim_{x\to +\infty}g(h(x))}$$

and then

$$\lim_{x\to +\infty}g(h(x)) =g(\lim_{x\to +\infty}h(x))$$

For a general discussion refer to:

• Analyzing limits problem Calculus (tell me where I'm wrong).

and other similar examples.

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For the solution we can avoid Taylor's series or l'Hospital as follows

$$\frac{e^x}{\left(1+\frac1x\right)^{x^2}} = e^{-\frac{\log\left(1+\frac1x\right)-x}{\frac1{x^2}}} \to e^{-\frac12}$$

using the method presented here.

Answer 2

(I've colored equals signs below as red when the argument is falty. Apologies to the color-blind.)

You've done this:

$$\lim_{x\to\infty} \frac{f(x)}{g(x)^x}\color{red}=\lim_{x\to\infty}\frac{f(x)}{(\lim_{y\to\infty} g(y))^x}$$

If you could do that kind of substitution, you could solve a lot of problems this way:

$$\lim_{x\to \infty}\left(1+\frac1x\right)^x\color{red}=\lim_{x\to\infty}\left(\lim_{y\to\infty} 1+\frac 1y\right)^x=\lim_x 1^x=1.$$

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The real question is why you'd think you could let one tiny part of the limit go to infinity first.

One way to see this is wrong is to compute:

$$\lim_{x\to\infty}\frac{g(x)^x}{(\lim_{y\to\infty} g(y))^x}$$

In your technique, this limit is $1.$

If $L=\lim_{y\to\infty} g(y),$ then let $h(x)=g(x)-L.$ $h(x)$ is the "error."

Since $g(x)=L+h(x),$ we can compute the limit:

$$\lim_{x\to\infty} \frac{g(x)^x}{L^x}=\lim\left(\frac{g(x)}{L}\right)^x=\lim\left (1+\frac{h(x)}{L}\right)^x.$$

All we really know is that $h(x)\to 0.$ So, for example, if $h(x)=\frac 1x,$ then the right side of the limit would be $e^{1/L}.$

In your case, you need an estimate for $h(x)=e-(1+1/x)^x.$ I'm not sure how to do that, but I'd bet, from the actual answer, that $h(x)=\frac{e}{2x}+o\left(\frac1x\right).$

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Another way to see the error is to take the first approach, but apply the second answer's logic. We'll just take the logarithms. Your second logic is:

$$\lim_{x\to\infty} (x-x^2\log(1+1/x))\color{red}=\lim_{x\to\infty} x-x\left(\lim_{y\to\infty} y\log(1+1/y)\right)=0.$$ since $y\log(1+1/y)\to 1.$

As you can see here, the common factor of $x$ is the problem. It is true that $1-x\log(1+1/x)\to 0,$ but it is not true that $x(1-x\log(1+1/x))\to 0.$