Question

Let $f(x)$ be defined as the definite integral from 0 to $x$ of the natural logarithm of $1 + t^2$, divided by $1 + t^4$, with respect to $t$. Now, define another function $g(x)$ as the value of $f(x)$ divided by $x$, for all $x$ not equal to 0. At $x = 0$, define $g(0)$ to be the limit of $g(x)$ as $x$ approaches 0. Is the function $g(x)$ differentiable at $x = 0$?

Answer 1

Yes

Answer 2

To determine if the function $g(x)$ is differentiable at $x=0$, we need to check if the limit $g'(0) = \lim_{h \to 0} \frac{g(0+h) - g(0)}{h}$ exists and is finite.

Step 1: Calculate $g(0)$

The function $g(x)$ is defined as $g(x) = \frac{f(x)}{x}$ for $x \neq 0$, and $g(0) = \lim_{x \to 0} g(x)$. We have $f(x) = \int_0^x \frac{\ln(1+t^2)}{1+t^4} dt$.

First, evaluate $f(0)$: $f(0) = \int_0^0 \frac{\ln(1+t^2)}{1+t^4} dt = 0$.

Now, calculate $g(0)$: $g(0) = \lim_{x \to 0} \frac{f(x)}{x}$. This is an indeterminate form $\frac{0}{0}$ since $f(0)=0$. We can use L'Hopital's Rule. By the Fundamental Theorem of Calculus, $f'(x) = \frac{d}{dx} \int_0^x \frac{\ln(1+t^2)}{1+t^4} dt = \frac{\ln(1+x^2)}{1+x^4}$. Applying L'Hopital's Rule: $g(0) = \lim_{x \to 0} \frac{f'(x)}{1} = \lim_{x \to 0} \frac{\ln(1+x^2)}{1+x^4} = \frac{\ln(1+0^2)}{1+0^4} = \frac{\ln(1)}{1} = \frac{0}{1} = 0$. So, $g(0) = 0$.

Step 2: Calculate $g'(0)$

The derivative $g'(0)$ is defined as: $g'(0) = \lim_{h \to 0} \frac{g(h) - g(0)}{h}$. Substitute $g(0) = 0$ and $g(h) = \frac{f(h)}{h}$ for $h \neq 0$: $g'(0) = \lim_{h \to 0} \frac{\frac{f(h)}{h} - 0}{h} = \lim_{h \to 0} \frac{f(h)}{h^2}$. This is again an indeterminate form $\frac{0}{0}$ since $f(0)=0$. We apply L'Hopital's Rule. $g'(0) = \lim_{h \to 0} \frac{f'(h)}{2h}$. Substitute $f'(h) = \frac{\ln(1+h^2)}{1+h^4}$: $g'(0) = \lim_{h \to 0} \frac{\frac{\ln(1+h^2)}{1+h^4}}{2h} = \lim_{h \to 0} \frac{\ln(1+h^2)}{2h(1+h^4)}$. This is still an indeterminate form $\frac{0}{0}$ since $\ln(1+0^2) = 0$ and $2(0)(1+0^4) = 0$. We apply L'Hopital's Rule again. Let $N(h) = \ln(1+h^2)$ and $D(h) = 2h(1+h^4) = 2h+2h^5$. Calculate their derivatives: $N'(h) = \frac{d}{dh}(\ln(1+h^2)) = \frac{1}{1+h^2} \cdot 2h = \frac{2h}{1+h^2}$. $D'(h) = \frac{d}{dh}(2h+2h^5) = 2+10h^4$. Applying L'Hopital's Rule for the second time: $g'(0) = \lim_{h \to 0} \frac{N'(h)}{D'(h)} = \lim_{h \to 0} \frac{\frac{2h}{1+h^2}}{2+10h^4}$. Substitute $h=0$: $g'(0) = \frac{\frac{2(0)}{1+0^2}}{2+10(0)^4} = \frac{0}{2} = 0$.

Since the limit $g'(0)$ exists and is finite (it is 0), the function $g(x)$ is differentiable at $x=0$.

Alternative method for $\lim_{h \to 0} \frac{\ln(1+h^2)}{2h(1+h^4)}$ using standard limits:

We know that $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$. Rewrite the limit: $\lim_{h \to 0} \frac{\ln(1+h^2)}{2h(1+h^4)} = \lim_{h \to 0} \left( \frac{\ln(1+h^2)}{h^2} \cdot \frac{h^2}{2h(1+h^4)} \right)$ $= \lim_{h \to 0} \left( \frac{\ln(1+h^2)}{h^2} \cdot \frac{h}{2(1+h^4)} \right)$. As $h \to 0$: $\lim_{h \to 0} \frac{\ln(1+h^2)}{h^2} = 1$ (by substituting $x=h^2$). $\lim_{h \to 0} \frac{h}{2(1+h^4)} = \frac{0}{2(1+0)} = 0$. Therefore, $g'(0) = 1 \cdot 0 = 0$.

Both methods confirm that $g'(0)=0$, which is a finite value.

The function $g(x)$ is differentiable at $x=0$.