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.

Since the original ratio is of the indeterminate form $\frac{0}{0}$, we use L'Hopital's rule:
$$
\lim_{x\to 0}\frac{2^x-1-x}{x^2}=\lim_{x\to 0}\frac{2^x\ln{2}-1}{2x}=\frac{\ln{2}-1}{0}=\infty
$$
Thus the limit does not exist at $x=0$ and the vertical asymptote makes total sense.
