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.
 A: There is a mistake in your steps.
Observe that, as $x \to 0$, by the Taylor expansion, one has
$$
e^u=1+u+\frac{u^2}2+O(u^3)
$$ giving

$$
\frac{2^x-1-x}{x^2}=\frac{e^{x \ln 2}-1-x}{x^2}=\frac{-1+\ln 2}{x}+\frac{1}{2}(\ln2)^2+O(x)
$$ 

which diverges as $x \to 0$.
A: Your mistake is that you used the limit laws wrongly. The limit of a sum is the sum of the limits if the two separate limits exist. Since you haven't proven that in your usage, and indeed it doesn't, the reasoning is flawed and hence does not contradict the actual fact that the limit does not exist.
Note also that you cannot use L'Hopital's rule to prove that a limit does not exist. Consider $f(x) = x^2 \sin(\frac{1}{x^2})$ and $g(x) = x$. Then $\frac{f(x)}{g(x)} \to 0$ as $x \to 0$ but $\frac{f'(x)}{g'(x)}$ is unbounded in any open neighbourhood of $0$, although $f(x),g(x) \to 0$ as $x \to 0$. So the non-existence of the limit after 'applying' L'Hopital's rule (ignoring the condition that the limit of the ratio of derivatives exists) does not imply non-existence of the original limit!
A: $\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}
$
A: 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.
