Answer
Let
[ f(x)=\frac{1}{\sqrt{1+e^{1/x}}},\qquad x\in\mathbb {R}\setminus{0}. ]
We have to find
[ \lim_{x\to 0}f(x),\qquad \lim_{x\to \pm \infty}f(x). ]
The behaviour of (f(x)) near (0) depends on the sign of (x).
If (x>0) and (x) is close to (0), then (1/x>0) and
(e^{1/x}\to +\infty). Hence
[ 1+e^{1/x}\to +\infty ,\qquad \sqrt{1+e^{1/x}}\to +\infty , ]
so
[ f(x)=\frac{1}{\sqrt{1+e^{1/x}}}\longrightarrow 0 . ]
Thus
[ \boxed{\displaystyle\lim_{x\to 0^{+}}f(x)=0 } . ]
If (x<0) and (|x|) is small, then (1/x<0) and
(e^{1/x}\to 0^{+}). Consequently
[ 1+e^{1/x}\to 1 ,\qquad \sqrt{1+e^{1/x}}\to 1 , ]
and
[ f(x)=\frac{1}{\sqrt{1+e^{1/x}}}\longrightarrow 1 . ]
Hence
[ \boxed{\displaystyle\lim_{x\to 0^{-}}f(x)=1 } . ]
Because the one‑sided limits are different, the two‑sided limit at (x=0) does not exist.
For large (|x|) the exponent (1/x) is close to (0).
Using the expansion (e^{t}=1+t+o(t)) as (t\to 0),
[ e^{1/x}=1+\frac{1}{x}+o!\left(\frac{1}{x}\right). ]
Hence
[ 1+e^{1/x}=2+\frac{1}{x}+o!\left(\frac{1}{x}\right) ]
and
[ \sqrt{1+e^{1/x}} =\sqrt{2},\sqrt{1+\frac{1}{2x}+o!\left(\frac{1}{x}\right)} =\sqrt{2}\left(1+\frac{1}{4x}+o!\left(\frac{1}{x}\right)\right). ]
Therefore
[ f(x)=\frac{1}{\sqrt{1+e^{1/x}}} =\frac{1}{\sqrt{2}}\left(1-\frac{1}{4x}+o!\left(\frac{1}{x}\right)\right) \xrightarrow[x\to\pm\infty]{} \frac{1}{\sqrt{2}} . ]
So
[ \boxed{\displaystyle\lim_{x\to\pm\infty}f(x)=\frac{1}{\sqrt{2}} } . ]
[ \begin{aligned} \lim_{x\to 0^{+}}f(x)&=0,\[2mm] \lim_{x\to 0^{-}}f(x)&=1,\[2mm] \lim_{x\to\pm\infty}f(x)&=\frac{1}{\sqrt{2}} . \end{aligned} ]
The function has a jump discontinuity at (x=0); it is continuous elsewhere. The limits at infinity show that the horizontal asymptote is (y=1/\sqrt{2}).
