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If $$\lim _{x\to 0}\left(1+x+\frac{f\left(x\right)}{x}\right)^{1/x} = e^3 \tag{1}$$ then evaluate $$\lim _{x\to 0}\left(1+\frac{f\left(x\right)}{x}\right)^{1/x} \tag{2}$$

Also, I want to know if $(1)$ is always equal to $e^3$? If not, at which $f$ the first equation is only true?

Blue
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Keneth Adrian Dagal
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  • You can use $\lim_{x\rightarrow 0^+} (1+rx)^{1/x}=e^r$ (which answers your final question) or $\lim_{n\rightarrow\infty} (1+\frac{r}{n})^n=e^r$. – Michael Mar 24 '20 at 11:37
  • Welcome to SE! Your claim is false for $f(x) = 0$, for example, as it is easy to see that you'd have $L:=\lim_{x \rightarrow 0}(1 + x + f(x)/x)^{1/x} = \lim_{x \rightarrow 0}(1 + x )^{1/x} = \lim_{z \rightarrow \infty}(1 + 1 / z)^z = e \ne e^3$. On the other hand, if $f(x)=2x^2$, then $L = \lim_{x \rightarrow 0}(1 + 3x)^{1/x} = \lim_{z \rightarrow \infty}(1 + 3 / z)^z = e^3$. – dohmatob Mar 24 '20 at 11:38

2 Answers2

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You don't need to figure out what $f$ is because frankly speaking the given condition does not guarantee a unique $f$. You can start by taking logs and note that this gives us $$\lim_{x\to 0}\frac{\log(1+x+f(x)/x)}{x}=3$$ which means that $x+f(x) /x\to 0$ and therefore we can rewrite the above limit as $$\lim_{x\to 0}\frac{\log(1+x+f(x)/x)}{x+f(x)/x}\cdot\frac{x+f(x)/x}{x}=3$$ The limit of first fraction is $1$ and hence the limit of second fraction is $3$. It follows that $f(x) /x^2\to 2$.

Next consider the fraction $$\frac{\log(1+f(x)/x)}{x}$$ which can be rewritten as $$\frac{\log(1+f(x)/x)}{f(x)/x}\cdot\frac{f(x)}{x^2}\to 1\cdot 2=2$$ and therefore the desired limit is $e^2$.

Paramanand Singh
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$\lim_{x\to 0}\left(1+x+\frac{f(x)}{x}\right)^{1/x}=\exp\left(\lim_{x\to 0}\frac{1}{x}\left(x+\frac{f(x)}{x}\right)\right)=e^3$. This implies that $f(x)\approx 2x^2$ for $x$ near $0$. This can be used directly to evaluate the other limit which comes out to be $e^2$.

Paras Khosla
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  • You should not use approximations like that. Limits are not about approximations rather they follow exact rules and one should use those rules – Paramanand Singh Mar 24 '20 at 13:52
  • @ParamanandSingh I was just trying to use the first term of the Taylor series of $f(x)$, isn't it actually in practice the same as using $\lim_{x\to 0}\frac{f(x)}{x^2}=2$? – Paras Khosla Mar 24 '20 at 15:28
  • I think it is better to be precise especially in calculus because there is already a lot of confusion created by trash calculus books who use all sorts of imprecise language and create oversimplification. While an experienced person can understand your answer fully for a novice this is a bit ambiguous. – Paramanand Singh Mar 24 '20 at 18:59