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the solution given in book is $1$ and by L'Hospital approach i do get one so what's the fault in method

Question:

$$\lim_{x\to 0} \frac{ e^x - e^{-x}-2\ln(1+x)}{x \sin x}$$

Work: $$\begin{aligned} \lim_{x\to 0} \frac{ e^x - e^{-x}-2\ln(1+x)}{x \sin x} & = \lim_{x\to 0} \frac{ x(\frac{e^x-1}{x}) +1 - \frac{1}{x(\frac{e^x-1}{x}) + 1}-2\frac{\ln(1+x)}{x}\cdot x}{x^2\cdot \frac{\sin x}{x} }\\& = \lim_{x\to 0} \frac{x + 1- \frac{1}{x+1} - 2x}{x^2}\\& = \lim_{x\to0}\frac{1 - x- \frac{1}{x+1}}{x^2}\\& = \lim_{x\to0} \frac{1 - x^2 - 1}{(x+1)x^2}\\& = \lim_{x\to 0}\frac{-1}{x+1}\\& = -1\end{aligned}$$

Why is this method wrong?

Sine of the Time
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zuck
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    http://math.stackexchange.com/help/notation – David K Jun 03 '23 at 05:12
  • You cannot insert limits of $\frac {e^{x}-1} x$ etc. – geetha290krm Jun 03 '23 at 05:15
  • i don't get it why ? – zuck Jun 03 '23 at 05:28
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    For now, I have edited your post using MathJax. But in the future, try to do it yourself and avoid adding images as your work. – Utkarsh Jun 03 '23 at 05:32
  • Why do you think your method is wrong, zuck? – Gerry Myerson Jun 03 '23 at 05:35
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    because the correct answer given is +1 and by L hospital rule also the answer is +1 – zuck Jun 03 '23 at 05:37
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    You cannot separate limit on summands, until you know both exists. – zkutch Jun 03 '23 at 05:42
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    The $\frac{\sin x}{x}$ replacement is fine because of the product rule for limits. Leaving that out, you're left with $\frac{xg(x)+1-\frac{1}{xg(x)+1}-2xh(x)}{x^2}$ where $g,h$ are functions which converge to $1$ as $x$ converges to $0$. By substituting $1$ in place of $g$ and $h$, you are effectively writing this as $\frac{xg(x)+1-\frac{1}{xg(x)+1} - 2xh(x)}{x+1-\frac{1}{x+1}-2x} \times \frac{x+1-\frac{1}{x+1}-2x}{x^2}$, and then claiming that because substituting $g$ and $h$ by $1$ is correct, the limit of the first fraction is $1$. But it's not. It actually equals $-1$. – Sarvesh Ravichandran Iyer Jun 03 '23 at 06:10
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    The reason for that is that even if you have four functions $t(x),s(x),u(x),v(x)$ such that $\frac{t(x)}{s(x)} = 1$ and $\frac{u(x)}{v(x)} = 1$, you cannot claim that $\frac{t(x)-u(x)}{s(x)-v(x)} = 1$ as well. In fact the latter limit may not even exist, as one can show with some examples. However, if you were to substitute $t(x)$ by $s(x)$ and $u(x)$ by $v(x)$, you would get $1$. That shows that substituting an expression by its limit is fallacious, especially when you see minus signs sticking around. Minus signs typically indicate that higher order Taylor coefficients are needed. – Sarvesh Ravichandran Iyer Jun 03 '23 at 06:13
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    You have replaced various parts ($(e^x-1)/x,(\log(1+x))/x,(\sin x) /x$) of the expression with their limits. Out of these the replacement of $(\sin x) /x$ with $1$ is valid while other replacements are invalid. I have discussed such replacements in detail in this answer https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh Jun 13 '23 at 04:24

2 Answers2

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$$\lim_{x\to0}\frac{e^x-e^{-x}-2\ln(1+x)}{x\sin{x}}$$ Substitute the quadratic Taylor approximation of each function $$\lim_{x\to0}\frac{\left(1+x+\frac{x^2}{2}\right)-\left(1-x+\frac{x^2}{2}\right)-2\left(x-\frac{x^2}{2}\right)}{x(x)}=\lim_{x\to0}\frac{x^2}{x^2}=1$$ Your method is wrong because the limit formulas that you used are linear approximations, which omit higher degree terms such as $x^2$

Amogh
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    Your result is fine but you should indicate the remainder term $O(x^3)$ in the expansion, as follows$$ \lim_{x\to0}\frac{\left(1+x+\frac{x^2}{2}\right)-\left(1-x+\frac{x^2}{2}\right)-2\left(x-\frac{x^2}{2}\right)+O(x^3)}{x(x+O(x^2))}=\lim_{x\to0}\frac{x^2+O(x^3)}{x^2+O(x^3)}=\lim_{x\to0}\frac{1+O(x)}{1+O(x)}=1$$ – user Jun 13 '23 at 06:57
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As noticed, we can't replace single part of the expression as you have done. Sometimes it works but the result is not guaranteed in general. Therefore this way to proceed should be avoided.

In some sense, the way you have used corresponds to a first order approximation, which should be performed as follows

  • $e^x=1+x+O(x^2)$
  • $\log(1+x)=x+O(x^2)$
  • $\sin x=x+O(x^2)$

to obtain

$$\frac{ e^x - e^{-x}-2\ln(1+x)}{x \sin x}=\frac{1+x-\frac1{1+x+O(x^2)}-2x+O(x^2)}{x^2+O(x^3)}=\frac{-x^2+O(x^2)}{x^2+O(x^3)}=\frac{-1+O(1)}{1+O(x)}$$

and here we see the issue hided behind your result: first order expansion doesn't suffice to obtain the result, indeed the term $O(1)$ in the numerator represents something which is not necessarly equal to zero.

This result explains also the importance to use the remainder term (big-O or little-o) when we use Taylor expansion in order to avoid wrong evaluations.

To obtain the correct limit, as indicated in the first given answer, we need to proceed with a second order expansion.

As an alternative we can also proceed without series expansion, using the results indicated here, as follows

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

$$=\frac x{\sin x}\left(\frac{e^x-1-x}{x^2}-\frac{e^{-x}-1+x}{x^2}-2\frac{\ln(1+x)-x}{x^2}\right)\to 1 \cdot \left(\frac12-\frac12-2\cdot -\frac12\right)=1$$

user
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