Help with the following limits 1) $$\lim_{x\to 1} \frac{x^{x}-x^{x^2}}{(1-x)^2} $$ . I tried solving it with $$ e^{glnf} $$
2) $$\lim_{x\to 0} ({\frac{(1+x)^\frac{1}{x}}{e}})^\frac{1}{x}$$
3) $$\lim_{x\to \infty}\frac{(x+\sqrt2)^{\sqrt2}-(x-\sqrt2)^{\sqrt2}}{x^{\sqrt2-1}} $$
I'd post my attempt but it didn't go nowhere and for the last one I don't even know where to start from.
 A: First lets factor $x^x$ out and as this limits to one the problem becomes
$$\lim\limits_{x\to 1}\frac{1-x^{x(x-1)}}{(x-1)^2}$$ Now let $x=y+1$ to get 
$$\lim\limits_{y\to 0}\frac{1-(y+1)^{y(y+1)}}{y^2}$$
or
$$\lim\limits_{y\to 0}\frac{1-e^{y(y+1)\ln(y+1)}}{y^2}$$
Now 
$$\frac{1-e^{y(y+1)\ln(y+1)}}{y^2}=\frac{1-e^{y(y+1)\ln(y+1)}}{y(y+1)\ln(y+1)}\frac{y(y+1)\ln(y+1)}{y^2}$$
$$=\frac{1-e^{y(y+1)\ln(y+1)}}{y(y+1)\ln(y+1)}(y+1)\frac{\ln(y+1)}{y}$$
And by elementary limits this is $(-1)(1)(1)=-1$.
A: For the second one:
$$\lim_{x\to0}\bigg(\frac{(1+x)^{\frac{1}{x}}}{e}\bigg)^{\frac{1}{x}} = \lim_{x\to0}\bigg[\bigg(1+\frac{(1+x)^{\frac{1}{x}}-e}{e}\bigg)^{\frac{e}{(1+x)^{\frac{1}{x}}-e}}\bigg]^{\frac{1}{x}\cdot\big(\frac{(1+x)^\frac{1}{x}-e}{e}\big)} = e^{\frac{1}{e}\cdot\lim_{x\to0}\frac{(1+x)^{\frac{1}{x}}-e}{x}} = l$$
Let's now focus on the upper limit:
$$\lim_{x\to0}e\cdot\frac{e^{\frac{\ln(1+x)}{x}-1}-1}{\frac{\ln(1+x)}{x}-1}\cdot\frac{\frac{\ln(1+x)}{x}-1}{x} = e\cdot\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$
Which is, after applying l'Hospital, equal to $\frac{-e}{2}$. Therefore, the final limit is :
$$l = e^{-\frac{1}{e}\cdot\frac{e}{2}}=e^{-1/2}=\frac{1}{\sqrt{e}}$$
A: $$\dfrac{x-x^{x^2}}{(1-x)^2}=x^x\cdot\dfrac{1-x^{x^2-x}}{(1-x)^2}$$
For the second part set $1-x=h$ to get $$\lim_{h\to0}\dfrac{1-(1-h)^{-h(1-h)}}{h^2}$$
Now use Binomial series, $$(1-h)^{-h(1-h)}=1-h^2(1-h)+O(h^3)$$
A: For the first one apply L'Hospitals rule twice so $$\lim _{ x\to 1 } \frac { x^{ x }-x^{ x^{ 2 } } }{ (1-x)^{ 2 } } =\lim _{ x\to 1 } \frac { { e }^{ x\ln { x }  }-{ e }^{ { x }^{ 2 }\ln { x }  } }{ (1-x)^{ 2 } } \overset { L'Hospital }{ = } -\lim _{ x\to 1 } \frac { { e }^{ x\ln { x }  }\left( \ln { x+1 }  \right) -{ e }^{ { x }^{ 2 }\ln { x }  }\left( 2x\ln { x } +x \right)  }{ 2\left( 1-x \right)  } \overset { L'Hospital }{ = } \\ =\lim _{ x\to 1 } \frac { { e }^{ x\ln { x }  }\left[ { \left( \ln { x+1 }  \right)  }^{ 2 }+\frac { 1 }{ x }  \right] -{ e }^{ { x }^{ 2 }\ln { x }  }\left[ { \left( 2x\ln { x } +x \right)  }^{ 2 }+2\ln { x+3 }  \right]  }{ 2 } =\frac { 2-4 }{ 2 } =-1$$
A: For the last one replace $\sqrt{2}$ by $n$ to simplify typing and let $x=1/t$ so that $t\to 0^{+}$. We then have $$L=\lim_{t\to 0^{+}}\frac{(1+nt)^{n}-(1-nt)^{n}}{t}$$ Now add subtract $1$ in numerator and use the standard limit $$\lim_{x\to a}\frac{x^r-a^r} {x-a} =ra^{r-1}$$ to get the answer as $L=n^{2}+n^{2}=4$.
The irrational number $\sqrt{2}$ has been used in the exponent only for intimidation and one usually beats this by replacing it with a generic symbol (like $n$ here).
