Limit exists theoritically,but graphically there is a vertical asymptote there.Why is it so. In finding the limit of $\lim_{x\to 0}\frac{2^x-1-x}{x^2}$
I used the substitution $x=2t$
$\lim_{x\to 0}\frac{2^x-1-x}{x^2}=\lim_{t\to 0}\frac{2^{2t}-1-2t}{4t^2}=\frac{1}{4}\lim_{t\to 0}\frac{2^{2t}-2\times2^t+2\times2^t+1-2-2t}{t^2}$
$=\frac{1}{4}\lim_{t\to 0}\frac{(2^{t}-1)^2+2\times2^t-2-2t}{t^2}$
$=\frac{1}{4}\lim_{t\to 0}\frac{(2^{t}-1)^2}{t^2}+\frac{1}{2}\lim_{t\to 0}\frac{2^t-1-t}{t^2}$
$\lim_{x\to 0}\frac{2^x-1-x}{x^2}=\frac{1}{2}(\ln2)^2$
But when i see the graph of the function,there is a vertical asymptote at $x=0$
And before $x=0$,function is approaching $\infty$ and after $x=0$,function is approaching $-\infty$.
That means limit should not exist,theoritically limit is coming but graphically limit does not exist.What is wrong?Why is it so?I do not understand.Please help me.

$\lim_{x\to 0}\frac{2^x-1-x}{x^2}
$
For small $x$,
$2^x
=e^{x \ln 2}
\approx 1+x \ln 2+O(x^2)
$
so
$\frac{2^x-1-x}{x^2}
=\frac{1+x \ln 2 +O(x^2)-1-x}{x^2}
=\frac{x (\ln 2-1) +O(x^2)}{x^2}
=\frac{(\ln 2-1) +O(x)}{x}
\to \infty
$
as
$x \to 0$.
In general,
$\lim_{x \to 0}\frac{a^x-1-x\ln a}{x^2}
$
exists
and is
$\begin{array}\\
\lim_{x \to 0}\frac{e^{x \ln a}-1-x\ln a}{x^2}
&=\lim_{x \to 0}\frac{1+(x \ln a)+(x \ln a)^2/2+O(x^3)-1-x\ln a}{x^2}\\
&=\lim_{x \to 0}\frac{(x \ln a)^2/2+O(x^3)}{x^2}\\
&=\lim_{x \to 0}( \ln a)^2/2+O(x)\\
&=( \ln a)^2/2\\
\end{array}
$
