# Calculate the following limit

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?

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 $\lim_{c \to 0} \ln c ( 1 - \sqrt{1 + c}) = 0$