How can i solve $\lim _{x\to +\infty } \left(\frac{3+x^{\alpha}}{2+x}\right)^{x}$ without de L'Hopital?

Question

I'm having some trouble with this limit:

$$\lim _{x\to +\infty } \left(\frac{3+x^{\alpha}}{2+x}\right)^{x}$$

The parameter $\alpha \in \mathbb{R}$. I've tried dividing the problem in three different cases: $\alpha = 1, \alpha < 1$ and $\alpha > 1$. For the first two the calculation looks pretty easy, as I, first, rewrite both numerator and denominator as the well known limit $(\theta/x + 1)^x \to e^\theta$: $$\frac{\left( 3 + x \right)^x}{\left( 2 + x \right)^x} = \frac{\left( 3/x + 1 \right)^x}{\left( 2/x + 1 \right)^x}\to \frac{e^3}{e^2} = e .$$

Then I see that for $\alpha > 1$: $$\left(\frac{3}{x^\alpha}+1\right)^x = \left(\frac{3}{x^\alpha} + 1 \right)^{x^\alpha x^{1-\alpha}}\to e^{3x^{1-\alpha}}\to1$$ and $$\left(\frac{2}{x^\alpha}+x^{1-\alpha}\right)^x \to 0.$$

The $\alpha < 1$ case, instead, gives me the indeterminate form $\frac\infty\infty$, which I'm unable to solve. This exercise is given in the "known limits" sections so I guess I can solve it without using de L'Hopital.

Answer 1

If $\alpha <1$,

$$\begin{align}\lim _{x\to \infty } \left(\frac{3+x^{\alpha}}{2+x}\right)^{x} &=\lim _{x\to \infty } \left(\frac{\frac 3x+x^{\alpha-1}}{\frac 2x+1}\right)^{x}\\ &=\lim _{x\to \infty } \left(x^x\right)^{\alpha -1}\\ &=\lim _{x\to \infty } \frac{1}{\left(x^x\right)^{1-\alpha}}=0. \end{align}$$

Answer 2

This step in your reasoning is not valid:

$$\left(\frac{3}{x^\alpha}+1\right)^x \to e^{x^{1-\alpha}}\to1$$

I assume your point is something like first rewrite $\left(\dfrac{3}{x^\alpha}+1\right)^x$ to $\left(\dfrac{3x^{1-\alpha}}{x}+1\right)^x$ (where did the 3 go, by the way?). However, the fact that you rely on next, $$ \Bigl(1+\frac{\theta}{x}\Bigr)^x \to e^\theta \qquad\text{when }x\to\infty $$ is only valid when $\theta$ is a constant! Here it actually depends on $x$, and it is not reliable reasoning to let only some of the $x$s in the limit go to $\infty$ first and then handle the rest later.

If we could do that, we could -- with the same justification -- decide to let the denominator in $\frac xx$ go to $\infty$ first and erroneously conclude $$ 1 = \frac xx = \frac1x\cdot x \to 0\cdot x \to 0 $$ as $x\to\infty$.

---

This is a case where trying to shoehorn your reasoning into purely symbolic rewrites makes things harder. In fact the $\alpha>1$ and $\alpha<1$ cases are both pretty easy:

• Suppose $\alpha>1$. Then $\frac{3+x^\alpha}{2+x}=\frac{3/x+x^{\alpha-1}}{2/x+1} \to \infty$ and therefore from a certain point onwards -- say, for $x>M$ -- we always have $\frac{3+x^\alpha}{2+x}>2$. Therefore when $x>\max(M,0)$ we have $$ \Bigl(\frac{3+x^\alpha}{2+x}\Bigr)^x > 2^x $$ which goes to $+\infty$.

• Suppose $\alpha<1$. Then the numerator of $\frac{3/x+x^{\alpha-1}}{2/x+1}$ goes to $0$ while the denominator goes to $1$, so from a certain point we always have $\frac{3+x^\alpha}{2+x}<\frac12$. Thus, from a certain point we'll have $$ \Bigl(\frac{3+x^\alpha}{2+x}\Bigr)^x < (1/2)^x $$ which just as easily goes to $0$.

The only complicated case is $\alpha=1$. Then we have $\frac{3+x^\alpha}{2+x}\to 1$, and neither of the above tricks will make progress. Then and only then is it time to be clever: $$ \Bigl(\frac{3+x}{2+x}\Bigr)^x = \Bigl(1+\frac{1}{2+x}\Bigr)^x = \Bigl(1+\frac{1}{2+x}\Bigr)^{2+x}\Bigl(1+\frac{1}{2+x}\Bigl)^{-2} $$ Now switching variable to $t=x+2$ we have $$ \lim_{x\to\infty}\Bigl(\frac{3+x}{2+x}\Bigr)^x = \lim_{t\to\infty} (1+\tfrac1t)^t(1+\tfrac1t)^{-2} = \big[\lim_{t\to\infty} (1+\tfrac1t)^t\big]\big[\lim_{t\to\infty}(1+\tfrac1t)^{-2}\big] = e\cdot 1 = e $$