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Compute the limit of the given function

Evaluate the limit.

    \[ \lim_{x \to 0} \left( \frac{1}{\log \left( x + \sqrt{1+x^2} \right)} - \frac{1}{\log(1+x)} \right). \]

( Note: There’s a small typo in Apostol which puts a parenthesis in the wrong place. The statement above is what I assume is meant since this evaluates to the answer given in the back of the book.)

To do this we’ll need to get the expression in the limit into the indeterminate form 0/0 and then apply L’Hopital’s rule twice. (Applying L’Hopital’s is going to be a challenge since the derivatives are going to get quite messy before we get anywhere.) We start by putting things over a common denominator,

    \[ \lim_{x \to 0} \left( \frac{1}{\log \left( x + \sqrt{1+x^2} \right)} - \frac{1}{\log (1+x)} \right) &= \lim_{x \to 0} \left( \frac{ \log(1+x) - \log \left( x + \sqrt{1+x^2} \right)}{\log \left( x + \sqrt{1+x^2} \right) \log (1+x)} \right). \]

Now we want to apply L’Hopital’s. First, we take the derivative of the numerator,

    \begin{align*}  D \left( \log (1+x) + \log \left( x + \sqrt{1+x^2} \right) \right) &= \frac{1}{1+x} + \frac{1}{x+\sqrt{1+x^2}} \cdot \left( 1 + \frac{x}{\sqrt{1+x^2}} \right) \\[9pt]  &= \frac{1}{1+x} + \frac{ 1 + \frac{x}{\sqrt{1+x^2}}}{x + \sqrt{1+x^2}} \\[9pt]  &= \frac{1}{1+x} + \frac{\sqrt{1+x^2} + x}{\sqrt{1+x^2} \left( x + \sqrt{1+x^2} \right)} \\[9pt]  &= \frac{1}{1+x} + \frac{1}{\sqrt{1+x^2}}. \end{align*}

Next, we take the derivative of the denominator,

    \begin{align*}  D \left( \log \left( 1 + \sqrt{1+x^2} \right) \log (1+x) \right) &= \frac{1 + \frac{x}{\sqrt{1+x^2}}}{1+\sqrt{1+x^2}} \cdot \log(1+x) + \frac{1}{1+x} \cdot \log \left( 1 + \sqrt{1+x^2} \right) \\[9pt]  &= \frac{ (1+x) \log (1+x) + \sqrt{1+x^2} \log \left( x + \sqrt{1+x^2} \right)}{(1+x)\sqrt{1+x^2}}. \end{align*}

So, now we use proceed with L’Hopital’s. (Keep in mind that L’Hopital’s is only valid if, once we get to the end, the limits of the derivatives we have taken actually exist, so right now we’re taking derivatives and hoping that the limits exist… once we have established that they do, then the whole process was valid.)

    \begin{align*}  \lim_{x \to 0} \left( \frac{ \log(1+x) - \log \left( x + \sqrt{1+x^2} \right)}{\log \left( x + \sqrt{1+x^2} \right) \log (1+x)} \right) &= \lim_{x \to 0} \frac{ \frac{1}{1+x} + \frac{1}{\sqrt{1+x^2}}}{ \frac{ (1+x) \log (1+x) + \sqrt{1+x^2} \log \left( x + \sqrt{1+x^2} \right)}{(1+x)\sqrt{1+x^2}} } \\[9pt]  &= \lim_{x \to 0}\frac{ (1+x)\sqrt{1+x^2} \left( \frac{1}{1+x} + \frac{1}{\sqrt{1+x^2}} \right)}{(1+x)\log(1+x) + \sqrt{1+x^2} \log \left(x + \sqrt{1+x^2} \right)} \\[9pt]  &= \lim_{x \to 0} \frac{ \sqrt{1+x^2} - (1+x)}{(1+x)\log(1+x) + \sqrt{1+x^2} \log \left(x + \sqrt{1+x^2} \right)}. \end{align*}

So, we again have the indeterminate form 0/0, and again we’ll try to apply L’Hopital’s. The derivative of the numerator is

    \begin{align*}  D \left( \sqrt{1+x^2} - (1+x) \right) &= \frac{x}{\sqrt{1+x^2}} -1 \\[9pt]  &= \frac{x - \sqrt{1+x^2}}{\sqrt{1+x^2}}. \end{align*}

The derivative of the denominator is

    \begin{align*}  D &\left( (1+x)\log(1+x) + \sqrt{1+x^2} \log \left(x + \sqrt{1+x^2} \right) \right) \\[9pt]  &= \log(1+x) + (1+x) \frac{1}{1+x} + \frac{x}{\sqrt{1+x^2}} \log \left( x + \sqrt{1+x^2} \right) + \sqrt{1+x^2} \frac{1 + \frac{x}{\sqrt{1+x^2}}}{x + \sqrt{1+x^2}} \\[9pt]  &= 2 + \log(1+x) + \frac{x \log \left( x + \sqrt{1+x^2} \right)}{\sqrt{1+x^2}}. \end{align*}

Putting these back into our evaluation of the limit,

    \begin{align*}  \lim_{x \to 0} \frac{ \sqrt{1+x^2} - (1+x)}{(1+x)\log(1+x) + \sqrt{1+x^2} \log \left(x + \sqrt{1+x^2} \right)} &= \lim_{x \to 0} \frac{ \frac{x - \sqrt{1+x^2}}{\sqrt{1+x^2}} }{2 + \log(1+x) + \frac{x \log \left( x + \sqrt{1+x^2} \right)}{\sqrt{1+x^2}}}. \end{align*}

But now, we have a quotient of continuous functions with non-zero denominator when x = 0 so the limit is the value of the function at x = 0. Evaluating we then have

    \[ \lim_{x \to 0} \left( \frac{1}{\log \left( x + \sqrt{1+x^2} \right)} - \frac{1}{\log(1+x)} \right) = -\frac{1}{2}.\]

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