Dilemma about value of limit $$
\lim_{x\to 0^{+}}\left[\left(1+\frac{1}{x}\right)^x+\left(\frac{1}{x}\right)^x+\left(\tan(x)\right)^{\frac{1}{x}}\right]$$
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
 A: 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
$$
1+1+0
$$
A: 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
$$\lim_{x\to 0^+}\left(1+\frac1x\right)^x=1$$

To evaluate the limit $\lim_{x\to 0^+}\left(\frac1x\right)^x$, we proceed as before.  We write simply
$$\begin{align}
\lim_{x\to 0^+}\left(\frac1x\right)^x&=\lim_{x\to 0^+}e^{-x\log(x)}\\\\&=1
\end{align}$$ 

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