Evaluate $\lim_{x\to 1^{+}}(\sqrt{x}-1)^{x^2+2x-3}$ using L'Hôpital I have solved before problems with L’Hopital’s Rule but this one is giving me a headache... Here it is:
$$\lim_{x\to 1^{+}}(\sqrt{x}-1)^{x^2+2x-3}$$
I know that first you need to $ \log$ it so you can get the $x^2+2x-3$ upfront and then you find the derivative till it is not longer $0/0$ or $\infty/\infty$, but I am doing all that and still can't find solution. If someone can solved it I would really appreciate it. 
Thank you
 A: If you take the log of it, it then becomes $\lim_{x\to1^+} (x^2+2x-3)\times \ln(\sqrt{x}-1)$.
Then you need to rewrite it as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ and then use the L'Hopital's Rule. 
So you can rewrite this as $\lim_{x\to1^+}\frac{x^2+2x-3}{\frac{1}{\ln(\sqrt{x}-1)}}$ or $\lim_{x\to1^+} \frac{\ln(\sqrt{x}-1)}{\frac{1}{x^2+2x-3}}$.
Then you can use L'Hopital's Rule on both numerator and denominator.
A: First simplify,
$$\lim_{x\to 1^{+}}(\sqrt{x}-1)^{x^2+2x-3}=\lim_{x\to 1^{+}}\left(\frac{x-1}{\sqrt{x}+1}\right)^{(x-1)(x+3)}
=\lim_{t\to0^+}\frac{(t^t)^4}{2^0}.$$
Then by L'Hospital, after taking the logarithm,
$$\lim_{t\to0^+}\ln(t^t)=\lim_{t\to0^+}\dfrac{\ln(t)}{\dfrac1t}=\lim_{t\to0^+}\frac{\dfrac1t}{-\dfrac1{t^2}}=0.$$
A: It's better to do some more work, before resorting to l'Hôpital's theorem
You're right in taking the logarithm, so you want to look at
$$
\lim_{x\to1^+}(x^2+2x-3)\log(\sqrt{x}-1)
$$
You can first note that $x^2+2x-3=(x-1)(x+3)$, so you can consider 
$$
\lim_{x\to1^+}(x-1)\log(\sqrt{x}-1)
$$
and you will reinsert the factor $x+3$ later. For this limit, you can do the substitution $\sqrt{x}-1=t$, so $x=(t+1)^2$ and the limit becomes
$$
\lim_{t\to0^+}(t+2)t\log t
$$
Now
$$
\lim_{t\to0^+}t\log t=\lim_{t\to0^+}\frac{\log t}{1/t}
\overset{*}{=}
\lim_{t\to0^+}\frac{1/t}{-1/t^2}=
\lim_{t\to0^+}(-t)=0
$$
(application of l'Hôpital marked with $*$).
Therefore
$$
\lim_{x\to1^+}(x-1)\log(\sqrt{x}-1)=
\lim_{t\to0^+}(t+2)t\log t
=\lim_{t\to0^+}(t+2)\cdot \lim_{t\to0^+}t\log t=0
$$
and so
$$
\lim_{x\to1^+}(x^2+2x-3)\log(\sqrt{x}-1)
=
\lim_{x\to1^+}(x+3)\cdot\lim_{x\to1^+}(x-1)\log(\sqrt{x}-1)
=0
$$
Finally, your original limit is
$$
\lim_{x\to1^+}(\sqrt{x}-1)^{x^2+2x-3}=e^0=1
$$

You could do it directly with the above substitution:
$$
\lim_{x\to1^+}(\sqrt{x}-1)^{x^2+2x-3}=
\lim_{t\to0^+}t^{t(t+2)((t+1)^2+3)}=
\lim_{t\to0^+}(t^t)^{(t+2)((t+1)^2+3)}=1^8=1
$$
because $\lim_{t\to0^+}t^t=1$.
A: Notice 
  $$(\sqrt{x} - 1)^{x^{2} + 2x - 3} = ((\sqrt{x} - 1)^{x-1})^{x + 3}.$$
Now 
  $$lim_{x \rightarrow 1^{+}} \ln((\sqrt{x} - 1)^{x-1}) = - lim_{x \rightarrow 1^{+}} \frac{\ln (\sqrt{x} - 1)}{(1-x)^{-1}} =^{L'H} lim_{x \rightarrow 1^{+}} (\frac{1}{2\sqrt{x}})(1-x)^{2}(1-\sqrt{x}) = 0,$$
so 
  $$lim_{x \rightarrow 1^{+}}(\sqrt{x} - 1)^{x^{2} + 2x - 3} = e^{0^{4}} = 1.$$
A: I think things simpify by converting the limit to one for $x\to0^+$:
$$
  \lim_{x\to 1^+}(\sqrt x -1)^{x^2+2x-3} 
= \lim_{x\to 0^+}(\sqrt{x+1}-1)^{(x+1)^2+2(x+1)-3}.
$$
Now
$$
  \lim_{x\to 0^+}(\sqrt{x+1}-1)^{x^2+4x}
=\exp\left(\lim_{x\to 0^+}\left((x+4)x\ln(\sqrt{x+1}-1)\right)\right)
$$
and we can estimate near $x=0^+$
$$
  x\ln(\sqrt{x+1}-1)=x\ln\left(x\,(\textstyle\frac12+o(x)\right)
  =x\bigl(\ln(x)-\ln2+o(x)\bigr),
$$
which, since $\lim_{x\to 0^+}x\ln(x)=0$, has limit $0$ at $x=0$.
Then the original limit becomes $\exp(4\times0)=1$.
In this approach I found no occasion to apply l’Hôpital’s Rule, though one might invoke it to prove that $\lim_{x\to 0^+}x\ln(x)=0$.
