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I need to solve $$ \lim_{x\to 0}\ (\sqrt {2x+1}\ -\ \sqrt[3]{1-3x})^{x}$$ Please note that I'm first year student and that this can be solved much simpler than in the answers. I tried doing $$\lim_{x\to 0} \ e^{x \cdot \ln\Bigl(\sqrt{2x+1}-1-\left(\sqrt[3]{1-3x}-1\right)\Bigr)}$$ then going with the limit inside the function like this

$$\exp\left\{\lim_{x\to0}x \cdot \ln\left[\lim_{x \to 0}\Bigl(\sqrt{2x+1}-1\Bigr) \cdot \lim_{x \to 0} \left(1- \frac{ \sqrt[3]{1-3x}-1\over x }{ \sqrt{2x+1}-1 \over x }\right)\right] \right\}$$

But problem is that although I can solve third limit this way, I get that second limit is 0, which makes that 0 is inside of $\ln$ and thus is incorrect attempt. Please help, I'm new here, I wan't to contribute back and this is from my university math exam.

Asaf Karagila
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dzaralica69
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    By the way, the reason your attempt does not work is that you are allowed to push the operator $\lim_{x \to x_0}$ inside the argument of a function $f$ only when $f$ is continuous $x_0$. In this case, the function $\ln$ is not continuous at $x_0 = 0$ because it is not even defined there. – giobrach Jun 29 '19 at 19:39
  • Thanks, I appreciate it. – dzaralica69 Jun 29 '19 at 19:48

4 Answers4

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Use equivalence: $$\bigl(\sqrt {2x+1} -\sqrt[3]{1-3x}\bigr)^{x}=\mathrm e^{x\ln(\sqrt {2x+1} - \sqrt[3]{1-3x})}. $$ Now by the binomial expansion at order $1$: $$(1+x)^\alpha=1+\alpha x+o(x),$$ one obtains, with $\alpha=\frac12$ and $\alpha=\frac13$, $$\sqrt {2x+1} -\sqrt[3]{1-3x}=(1+x+o(x))-(1-x+o(x))=2x+o(x),$$ so that $ \;\sqrt {2x+1} - \sqrt[3]{1-3x}\sim_0 2x$, and ultimately $$x\ln(\sqrt {2x+1} - \sqrt[3]{1-3x})\sim_0 x\ln (2x)=x\ln 2+x\ln x,$$ which tends to $0$ when $x$ tends to $0$, so that $$\lim_{x\to 0}\bigl(\sqrt {2x+1} -\sqrt[3]{1-3x}\bigr)^{x}=\mathrm e^0=1.$$

Bernard
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We shall only try to find $\lim_{x\to 0^+}$ because for negative $x$ near $0$, the power is not defined (for reasons given below).

Working without little-o stuff:

Note that $$(1+x)^2=1+2x+x^2\ge 1+2x$$ for all $x$ and of course $1<1+2x$ for all $x>0$. We conclude that $$1 <\sqrt{1+2x}\le 1+x\qquad\text{for }x>0.$$

Similarly, $$ (1-x)^3=1-3x+3x^2-x^3>1-3x\qquad \text{for }x<3$$ and $$(1-2x)^3=1-6x+12x^2-8x^3<1-3x-3x(1-4x)<1-3x \qquad \text{for }0<x<\frac14,$$ hence $$1-2x<\sqrt[3]{1-3x}<1-x\qquad\text{for }0<x<\frac14 $$ and so $$ x<\sqrt{1+2x}-\sqrt[3]{1-3x}<4x\qquad\text{for }0<x<\frac14.$$ (One can find similar bounds for negative $x$, showing that $\sqrt{1+2x}-\sqrt[3]{1-3x}\sim x<0$, and therefore $(\sqrt{1+2x}-\sqrt[3]{1-3x})^x$ is undefined for negative $x$ near $0$)

If we already know that $\lim_{x\to 0^+}x^x=1$, it follows that $(\sqrt{1+2x}-\sqrt[3]{1-3x})^x$ is squeezed between $x^x$ and $4^x\cdot x^x$ and, as $\lim_{x\to 0^+}4^x=1$, therefore also $$ \lim_{x\to0^+}(\sqrt{1+2x}-\sqrt[3]{1-3x})^x=\lim_{x\to0^+}x^x=1.$$

Why is $\lim_{x\to 0^+}x^x=1$?

Perhaps the most important inequality about the exponential is $$ e^t\ge 1+t\qquad \text{for all }t\in\Bbb R.$$ Therefore, for $t>0$, $$ e^t=(e^{t/2})^2\ge(1+\tfrac t2)^2=1+t+\frac14t^2>\frac14t^2.$$ It follows that $$0\le \lim_{t\to +\infty}\frac{t}{e^t}\le \lim_{t\to +\infty}\frac{t}{\frac14t^2}=0.$$ With $x=e^{-t}$ (i.e., $t=-\ln x$), this becomes $$\lim_{x\to 0^+} x\ln x=0$$ and therefore $$\lim_{x\to0^+} x^x=\lim_{x\to 0^+} e^{x\ln x}=e^{\lim_{x\to 0^+} x\ln x} =e^0=1.$$

Hagen von Eitzen
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$$\left(\sqrt{1+2x}-\sqrt[3]{1-3x}\right)^x=(1+x+o(x)-1+x+o(x))^x=2^xx^x(1+o(1))^x.$$

$2^xx^x \rightarrow 1$, $(1+o(1))^x=e^{xo(1)}=e^{o(1)}=1+o(1)$, thus the limit is $1$.

Aphelli
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Hint:

As lcm$(2,3)=6$

We have $$\dfrac{\lim_{x\to0}((1+2x)^3-(1-3x)^2)^x}{\lim_...(\sum_{r=0}^5((1-2x)^{r/2}(1-3x)^{(6-r)/3})^x}$$

The denominator tends to $1$

The numerator $$=\lim_...(12x+3x^2+8x^3)^x=\lim_..(12x)^x(1+x/4+2x^2/3)^x=1$$

lab bhattacharjee
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