Is the formula for the form $1^{\infty}$ that is $$\lim\limits_{x\to a} (1+f(x))^{g(x)}=e^{\lim\limits_{x\to a}f(x)g(x)}$$ a standard formula? Does it have any special name? Note:In the above formula $f(x)\to 0$ as $x\to a$ and $g(x)\to\infty$ as $x\to a$

I'm asking because I'm not sure whether it will be allowed in a subjective maths exam(it sometimes greatly simplifies calculations).Thanks.

  • $\begingroup$ I'm a bit confused as to what you're asking, but the constant functions $f = g = 1$ provide an obvious counter example if I have interpreted it correctly! $\endgroup$ – Zestylemonzi Oct 2 '15 at 11:13
  • $\begingroup$ @Zestylemonzi I guess you did'nt understand my question.I've edited it.Check now. $\endgroup$ – user220382 Oct 2 '15 at 11:17
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    $\begingroup$ Personally I would not accept it on my exams without explanations (very easy to explain via $\frac{\ln(1+f(x))}{f(x)}\to 1$). $\endgroup$ – A.Γ. Oct 2 '15 at 11:45
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    $\begingroup$ Fully agree with @A.G. It should be quite obvious that this is not a standard formula for evaluation of limits, but a cheap technique mentioned in many low quality calculus books to convince the student that learning calculus is merely learning a bag of tricks. $\endgroup$ – Paramanand Singh Oct 4 '15 at 5:26
  • $\begingroup$ Can you explain why it is not correct?@ParamanandSingh $\endgroup$ – cmi Aug 13 '18 at 7:41

One of the logarithmic inequalities says: $$\frac{x}{1+x} < \ln (1 + x) < x, \forall x > -1$$

Because $\lim\limits_{x\rightarrow a} f(x)=0$, then $f(x)>-1$, for all $x$ in some vicinity of $a$. So $$\frac{f(x)}{1+f(x)} < \ln (1 + f(x)) < f(x)$$ from all $x$ in some vicinity of $a$. As a result $$\lim_{x \rightarrow a} \left ( 1 + f(x) \right )^{\frac{1}{f(x)}}=e$$

Alternatively (let's limit to the case when $g(x)>0$) $$\frac{f(x) \cdot g(x)}{1+f(x)} < g(x) \cdot \ln (1 + f(x)) < f(x)\cdot g(x)$$ ($e^{x}$ is ascending $\forall x$) $$e^{\frac{f(x) \cdot g(x)}{1+f(x)}} < (1 + f(x))^{g(x)} < e^{f(x)\cdot g(x)}$$ Or $$\frac{e^{\frac{f(x) \cdot g(x)}{1+f(x)}}}{e^{f(x)\cdot g(x)}} < \frac{(1 + f(x))^{g(x)}}{e^{f(x)\cdot g(x)}}<1$$ Or $$0 < \frac{1}{e^{f(x) \cdot g(x) \cdot \frac{f(x)}{1+f(x)}}} < \frac{(1 + f(x))^{g(x)}}{e^{f(x)\cdot g(x)}}<1 \tag{*}$$ Now a little summary:

  • if $g(x)$ is bounded (at least in some vicinity of $a$), then $\mathbf{(*)}$ is asymptotic to $1$ when $x\rightarrow a$.
  • same is true if $g(x)$ happens to be continuous in a vicinity of $a$.

Not totally sure about the $\lim\limits_{x\rightarrow a} g(x)=\infty $ case. I would say, it depends.


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