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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.

enter image description here 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.

Priyanka
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    Well, you might try writing it as $\lim_{x \to 0} \frac{\lim_{y \to 0} \frac{2^y-1}{y}-1}{x}$. One way or another you realize that the inner limit is $\ln(2)$, and since $\ln(2) \neq 1$ the whole thing blows up. This is not really different from L'Hospital's rule but it is not literally L'Hospital's rule. – Ian Jun 23 '16 at 17:36
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    I bet that $-x$ should be replaced by $-x\log 2$ to have a finite limit. – Jack D'Aurizio Jun 23 '16 at 18:04
  • The answer according to our text book is $\frac{(\ln2)^2}{2}$ – Priyanka Jun 23 '16 at 18:05
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    Then your book is wrong, or (probably) there is some typo. – Jack D'Aurizio Jun 23 '16 at 18:05
  • But by using L'Hopital's rule you get the answer as given in our text book. – Priyanka Jun 23 '16 at 18:12
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    With $-x$ and not $-x\ln(2)$, no you don't. – Ian Jun 23 '16 at 18:19
  • @JackD'Aurizio But by using L'Hopital's rule you get the answer as given in our text book. – Priyanka Jun 23 '16 at 18:19
  • @Priyanka: then you are assuming $\frac{d}{dx}2^x=2^x$, but that is not true. – Jack D'Aurizio Jun 23 '16 at 18:22
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    @JackD'Aurizio, actually I think the OP's error (and possibly the book's error as well) lies in not checking the $0/0$ condition that's needed to justify a second round of L'Hopital. – Barry Cipra Jun 23 '16 at 18:36
  • @JackD'Aurizio , I have attached an image of the solution given by the text book. I am not able to figure out the mistake in the solution given by the text book. – Priyanka Jun 23 '16 at 18:56
  • @Priyanka The mistake comes from assuming you can write $L = \lim_{x\rightarrow0} \frac{2^x - x - 1}{x^2}$. No such $L$ exists because the limit doesn't exist. If the limit existed, then this would work. – Adam Francey Jun 23 '16 at 20:17
  • Either that book has a huge mistake or else it is giving an excellent example that you cannot apply arithmetic of limits, as done in the step from line three to line four, if you don't prove before hand that the limits of both expressions exist finitely . – DonAntonio Jun 23 '16 at 22:06
  • @Joanpemo. I agree. It was a huge mistake in the book. Please see the "Note" I added to the original problem statement. – Priyanka Jun 24 '16 at 03:12
  • +1 for this very good question. It shows that evaluating limits requires us to work precisely under the constraints of the valid theorems on limits. The fundamental mistake in the solution is assuming the existence of limit $L$. You can't and shouldn't assume a limit exists without evaluating it. – Paramanand Singh Jun 24 '16 at 03:59
  • If you are preparing for jee you can join us on http://chat.stackexchange.com/rooms/48921/discussions-in-pcm-and-anything-else –  Nov 23 '16 at 17:08

2 Answers2

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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.)

Barry Cipra
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    +1 for the Added explanation. It shows why such techniques which assume existence of limit (to be evaluated) are wrong. I have seen use of this technique here on MSE to evaluate $\lim_{x \to 0}(x - \sin x)/x^{3}$, but here the limit actually exists and the technique seems to work OK. – Paramanand Singh Jun 24 '16 at 04:02
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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

Stefan4024
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  • I think what you did is wrong, because lim(f.g)= lim(f).lim(g) only when both the limits exist. In your case g(x) = 1/x. Hence, lim(g) doesn't exist. So you cant do that. However , the answer according to our textbook is [(ln2)^2]/2 – Priyanka Jun 23 '16 at 17:53
  • @Priyanka The limit doesn't converge for sure. In order to make yourself sure plug a really really small number, like 0.000001 and check the result. Anyway when we can split the limit in one divergent and one convergent and we get a determinate form we can deduce that it's divergent. – Stefan4024 Jun 23 '16 at 18:10
  • using L'Hopital's rule u get a finite limit (which is same as the answer as given in our text book). – Priyanka Jun 23 '16 at 18:18
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    @Priyanka The problem is that after applying L'Hopital Rule once you get: $$\lim_{x \to 0} \frac{2^x \cdot \ln(2) - 1}{2x} = -\infty$$ You can't apply L'Hopital Rule again, as the fraction isn't one of the forms suitable for L'Hopital Rule – Stefan4024 Jun 23 '16 at 18:20
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    @Stefan4024 $\lim_{x \to 0^+}$ is what you wrote, I think $\lim_{x \to 0^-}$ is $+\infty$. – Ian Jun 24 '16 at 01:16
  • @Ian Yeah you're right from the LHS the limit approaches $\infty$ and $-\infty$ from the RHS – Stefan4024 Jun 24 '16 at 01:32