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
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 $\lim_{t\to0^+}t^t=1$.
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.
\ln instead of ln to properly format the $\ln$ function.
– Erick Wong
May 30 '16 at 17:47
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,
$$\lim_{t\to0^+}\ln(t^t)=\lim_{t\to0^+}\dfrac{\ln(t)}{\dfrac1t}=\lim_{t\to0^+}\frac{\dfrac1t}{-\dfrac1{t^2}}=0.$$
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 $$lim_{x \rightarrow 1^{+}}(\sqrt{x} - 1)^{x^{2} + 2x - 3} = e^{0^{4}} = 1.$$
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 $\lim_{x\to 0^+}x\ln(x)=0$.
