Fallacy limit problem - Where is the mistake? This problem comes from our text book. 
Evaluate 
$$
\lim\limits_{x \to 0}\frac{2^x-1-x}{x^2}
$$
without using either L'Hopital's rule or Taylor series.
The picture below shows the solution given by the text book. 

Many of the people who responded to this question say that the limit diverges. But according to the solution (see the attached image) given by the text book the limit is  $ \frac{(\ln2)^2}{2} $. Is there an error in the solution given in the attached image?
Note: This problem ( and the solution) comes from Cengage, a long time favourite of the students preparing for IIT JEE  Exam (IITs are India's most prestigious engineering institutes.)  It is suprising that no student/teacher has found this mistake all these days. However, once posted on Stackexchange, the fallacy was resolved in a very short time.
Thanks to all the people who responded to this problem. Now the mistake in the book's solution is found.
 A: Consider the limit when the denominator is $x$ instead an use the definition of derivatives for $f(x) = 2^x$.
$$\lim_{x \to 0} \frac{2^x - 1 - x}{x} = \lim_{x \to 0} \frac{2^x - 2^0}{x - 0} + \lim_{x \to 0} - \frac{x}{x} = \lim_{x \to 0} 2^x \cdot \ln(2) - 1 = \ln(2) - 1 \not = 0$$
Now when we divide by some number very close to zero we're bound to get infinity or minus infinity
A: Edit:  The OP has added an image from the book, giving an (incorrect!) evaluation of the limit.  I'm leaving my initial answer as is, and adding an explanation of where the OP's book went wrong at the end.
The OP, in comments, indicates the belief that L'Hopital's rule gives the book's answer.  But it only does so if you apply it blindly:
$${2^x-1-x\over x^2}\to{\ln2\cdot2^x-1\over2x}\to{(\ln2)^22^x\over2}\to{(\ln2)^2\over2}$$
where the first two arrows indicate L'Hopital steps and the last arrow is the evaluation of the trivial limit $2^x\to2^0=1$.
The error in this is that you can only apply L'Hopital's rule when both numerator and denominator tend to $0$.  So the first arrow is OK, since $2^0-1-0=1-1-0=0$.  But the second arrow is not, since $\ln2\cdot2^0-1=\ln2-1\not=0$.  
Added explanation.  The book's derivation would be OK if the limit $L$ exists.  The problem is, the limit does not exist (as Stefan4024's answer shows).  This is actually quite a nice example of what can go wrong if you don't pay close attention to the hypotheses that underlie the theorems for evaluating limits.  (In fact, what might be interesting here is to explain how two seemingly quite different incorrect derivations give the same wrong answer.)
