Xjyuk

I need to solve $$ lim_xto 0 (sqrt 2x+1 - sqrt[3]1-3x)^x$$ Please note that I'm first year student and that this can be solved much simpler than in the answers. I tried doing $$lim_xto 0 e^x cdot lnBigl(sqrt2x+1-1-left(sqrt[3]1-3x-1right)Bigr)$$ then going with the limit inside the function like this

$$expleftlim_xto0x cdot
lnleft[lim_x to 0Bigl(sqrt2x+1-1Bigr) cdot
lim_x to 0 left(1-
frac sqrt[3]1-3x-1over x sqrt2x+1-1 over x right)right] right$$

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.

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

Working without little-o stuff:

Note that $$(1+x)^2=1+2x+x^2ge 1+2x$$ for all $x$
and of course $1<1+2x$ for all $x>0$.
We conclude that
$$1 <sqrt1+2xle 1+xqquadtextfor x>0.$$

Similarly,
$$ (1-x)^3=1-3x+3x^2-x^3>1-3xqquad textfor x<3$$
and
$$(1-2x)^3=1-6x+12x^2-8x^3<1-3x-3x(1-4x)<1-3x qquad textfor 0<x<frac14,$$
hence
$$1-2x<sqrt[3]1-3x<1-xqquadtextfor 0<x<frac14 $$
and so
$$ x<sqrt1+2x-sqrt[3]1-3x<4xqquadtextfor 0<x<frac14.$$
(One can find similar bounds for negative $x$, showing that $sqrt1+2x-sqrt[3]1-3xsim x<0$, and therefore $(sqrt1+2x-sqrt[3]1-3x)^x$ is undefined for negative $x$ near $0$)

If we already know that $lim_xto 0^+x^x=1$, it follows that $(sqrt1+2x-sqrt[3]1-3x)^x$ is squeezed between $x^x$ and $4^xcdot x^x$ and, as $lim_xto 0^+4^x=1$, therefore also
$$ lim_xto0^+(sqrt1+2x-sqrt[3]1-3x)^x=lim_xto0^+x^x=1.$$

Why is $lim_xto 0^+x^x=1$?

Perhaps the most important inequality about the exponential is
$$ e^tge 1+tqquad textfor all tinBbb R.$$
Therefore, for $t>0$,
$$ e^t=(e^t/2)^2ge(1+tfrac t2)^2=1+t+frac14t^2>frac14t^2.$$
It follows that
$$0le lim_tto +inftyfracte^tle lim_tto +inftyfractfrac14t^2=0.$$
With $x=e^-t$ (i.e., $t=-ln x$), this becomes
$$lim_xto 0^+ xln x=0$$
and therefore
$$lim_xto0^+ x^x=lim_xto 0^+ e^xln x=e^lim_xto 0^+ xln x =e^0=1.$$

Use equivalence:
$$bigl(sqrt 2x+1 -sqrt[3]1-3xbigr)^x=mathrm e^xln(sqrt 2x+1 - sqrt[3]1-3x). $$
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]1-3xsim_0 2x$, and ultimately
$$xln(sqrt 2x+1 - sqrt[3]1-3x)sim_0 xln (2x)=xln 2+xln x,$$
which tends to $0$ when $x$ tends to $0$, so that
$$lim_xto 0bigl(sqrt 2x+1 -sqrt[3]1-3xbigr)^x=mathrm e^0=1.$$

$$left(sqrt1+2x-sqrt[3]1-3xright)^x=(1+x+o(x)-1+x+o(x))^x=2^xx^x(1+o(1))^x.$$

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

Hint:

As lcm$(2,3)=6$

We have $$dfraclim_xto0((1+2x)^3-(1-3x)^2)^xlim_...(sum_r=0^5((1-2x)^r/2(1-3x)^(6-r)/3)^x$$

The denominator tends to $1$

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