Dilemma about value of limit

Question

$$ \lim_{x\to 0^{+}}\left[\left(1+\frac{1}{x}\right)^x+\left(\frac{1}{x}\right)^x+\left(\tan(x)\right)^{\frac{1}{x}}\right]$$

Attempt: I used $\tan(x)\approx x$ also $(1+n)^{1/n}=e$ so I let $x=0+h$ and lim changes to lim h tending to $0$. But that gives me $e+\infty+h^{1/h}$. While the answer is an integer between $0-9$ . Where is my mistake?.Thanks

Answer 1

Since you seem to be losing yourself among infinitesimals, I can suggest a safer way; when exponentials are involved, passing to the logarithm is often simpler.

Besides, the substitution $x=0+h$ is of no consequence.

Consider $$ \lim_{x\to0^+}\log\left(\left(1+\frac{1}{x}\right)^{x}\right)= \lim_{x\to0^+}x\log\left(1+\frac{1}{x}\right)= \lim_{t\to\infty}\frac{\log(1+t)}{t}=0 $$ Moreover $$ \lim_{x\to0^+}\log\bigl((1/x)^x\bigr)= \lim_{x\to0^+}-x\log x=0 $$ Similarly, $$ \lim_{x\to0^+}\log\bigl((\tan x)^{1/x}\bigr)= \lim_{x\to0^+}\frac{\log\tan x}{x}=-\infty $$ Thus your limit is $$ 1+1+0 $$

Answer 2

Note that we can write

$$\begin{align} \left(1+\frac1x\right)^x&=e^{x\log\left(1+\frac1x\right)}\\\\ &=e^{x\left(\log(1+x)-\log(x)\right)}\\\\ &=e^{-x\log(x)}e^{x\log(1+x)} \end{align}$$

Since $\lim_{x\to 0^+}x\log(x)=0$, we find using the continuity of the exponential function that

$$\lim_{x\to 0^+}\left(1+\frac1x\right)^x=1$$

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To evaluate the limit $\lim_{x\to 0^+}\left(\frac1x\right)^x$, we proceed as before. We write simply

$$\begin{align} \lim_{x\to 0^+}\left(\frac1x\right)^x&=\lim_{x\to 0^+}e^{-x\log(x)}\\\\&=1 \end{align}$$

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To evaluate the limit $\lim_{x\to 0^+}\left(\tan (x)\right)^{1/x}$, we use the analogous approach and write

$$\lim_{x\to 0^+}\left(\tan (x)\right)^{1/x}=\lim_{x\to 0^+}e^{\frac1x \log(\tan(x))}$$

Note that $1/x\to \infty$ while $\log(\tan(x))\to -\infty$. Thus, the product $\frac1x \log(\tan(x)) \to -\infty$ and we find that

$$\lim_{x\to 0^+}\left(\tan (x)\right)^{1/x}=0$$

Answer 3

Note first of all that $(\tan x)^{1/x}\to0^\infty=0$ as $x\to0^+$, so the trickiest-looking part of the limit is actually the easiest. More formally, $0\le\tan x\le1$ for $0\le x\le\pi/4$ and $1/x\ge4/\pi\gt1$ for $0\lt x\le\pi/4$, so

$$0\le(\tan x)^{1/x}\le\tan x\quad\text{for }0\lt x\le{\pi\over4}$$

and thus $(\tan x)^{1/x}\to0$ as $x\to0^+$ by the squeeze theorem.

As for the rest of the limit, it's convenient to write

$$\left(1+{1\over x}\right)^x+\left(1\over x\right)^x={(x+1)^x+1\over x^x}$$

and then "distribute" the limit:

$$\lim_{x\to0^+}\left[\left(1+{1\over x}\right)^x+\left(1\over x\right)^x\right]={\lim_{x\to0^+}((x+1)^x+1)\over\lim_{x\to0^+}(x^x)}={2\over\lim_{x\to0^+}(x^x)}$$

So all that remains is to show that $\lim_{x\to0^+}(x^x)=1$, which is equivalent to showing

$$\lim_{x\to0^+}x\ln x=0$$

Computing this limit without using L'Hopital requires additional steps. I will omit them here, unless there is a request for them in comments. (The other answers, I see, took this limit for granted.)