I need to evaluate the following limit, however, in doing so, I let $\sqrt{1+c}=1$ which I came to undertand that it's not valid. My procedure was the following: $$\lim_{c\to0}\left(-\ln(c)\sqrt{1+c}-\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right)\right)$$ $$=\lim_{c\to0}\ln\left(\frac{\left(1-\sqrt{1+c}\right)\left(1+\sqrt2\right)}{c\left(1+\sqrt{1+c}\right)\left(1-\sqrt2\right)}\right)=\ln\left(\lim_{c\to0}\frac{\left(1-\sqrt{1+c}\right)\left(1+\sqrt2\right)}{c\left(1+\sqrt{1+c}\right)\left(1-\sqrt2\right)}\right)$$ $$$$We can use L'Hôspital's rule on the fraction to obtain the simplified version: $$\ln\left(\lim_{c\to0}\left(\frac{3+2\sqrt2}{2+3c+2\sqrt{1+c}}\right)\right)=\boxed{\ln\left(\frac{3+2\sqrt2}4\right)}$$ The answer that I obtained matches with the answer of Wolfram Alpha even though there is a mistake. If I don't let $\sqrt{1+c}=1$ the limit becomes really difficult and I don't know how to solve it. Does someone have any ideas on how to solve it? And why is my answer correct even though the procedure is not?
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You can write this as
$$-\ln(c)\sqrt{1+c}-\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right) \\ = -\ln(c)\sqrt{1+c} + \ln c- \ln c -\ln\left(\frac{1+\sqrt{1+c}}{1-\sqrt{1+c}}\right)+\ln\left(\frac{1+\sqrt2}{1-\sqrt2}\right) \\ = \ln c ( 1 - \sqrt{1 + c}) + A(c) $$
where
$$A(c) = \ln\left(\frac{\left(1-\sqrt{1+c}\right)\left(1+\sqrt2\right)}{c\left(1+\sqrt{1+c}\right)\left(1-\sqrt2\right)}\right).$$
As you have shown
$$\lim_{c \to 0} A(c) = \ln\left(\frac{3+2\sqrt2}4\right)$$
You will get the correct limit for the original expression now because $\lim_{c \to 0} \ln c ( 1 - \sqrt{1 + c}) = 0$