How can one evaluate this limit? $$\lim_{x\to0} \big((1+x)^x-1\big)^x$$ I've already tried writing that using $f(x) = e^{\ln f(x)}$, but I don't dare going any further with this approach.
Evaluate $\lim_{x\to0} \big((1+x)^x-1\big)^x$
3
$\begingroup$
limits
exponential-function
2 Answers
2
Binomial expansion:
$$(1+x)^x=1+x^2+\frac12(x-1)x^3+\mathcal O(x^4)$$
Thus, we have
$$(1+x)^x-1=x^2+\frac12(x-1)x^3+\mathcal O(x^4)$$
Finally:
$$(x^2+\frac12(x-1)x^3+\mathcal O(x^4))^x=(x^x)^2(1+\frac12(x-1)x+\mathcal O(x^2))^x\\\to1\times1=1$$
Since $\lim_{x\to0}x^x=1$ and direct substitution into the second product.
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0Can you do without binomial too? Like by making function as ${e}^{log_e (f(x))} $ ? – 2017-02-01
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0@Ramanujan not easily... – 2017-02-01
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$\lim_{x\rightarrow 0}((1+x)^x-1)^x = \lim_{x\rightarrow 0}\bigg(e^{x\ln(1+x)}-1\bigg)^x$
$\displaystyle = \lim_{x\rightarrow 0}\bigg(1+\frac{x\ln(1+x)}{1!}+\frac{x^2\ln^2(1+x)}{2!}+\cdots \infty - 1\bigg)^x $ $\displaystyle = \lim_{x\rightarrow 0}x^x(\ln(1+x))^x\bigg(1+\ln(1+x)+\frac{\ln(1+x)}{2!}+\cdots \cdots \bigg)^{x} = \lim_{x\rightarrow 0}x^x(\ln(1+x))^x$
$\displaystyle \lim_{x\rightarrow 0}x^x \times \lim_{x\rightarrow 0}(\ln(1+x))^x = 1\times \lim_{x\rightarrow 0}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots \infty\right)^x$
$\displaystyle = \lim_{x\rightarrow 0}x^x\times \lim_{x\rightarrow 0}\left(1-\frac{x}{2}+\frac{x^2}{3}+\cdots \right)^x = 1\times 1 = 1$