Proving that length of a curve is $\infty$

Question

Let $f$ be a differentiable and continious function in $(0,1]$ and $lim_{x\to 0^+}f(x)=\infty$. Prove that the length of the curve on (0,1] is $\infty$. Steps I tried: $L=\int_0^1 \sqrt{1+f\prime(x)^2}dx\geq\int_0^1\sqrt{f\prime(x)^2}=\int_0^1f\prime(x)=f(1)-lim_{x\to0^+}f(x)=-\infty$ (and I had to prove $\int\geq\infty$ and not $-\infty$).

Answer 1

I think your question is wrong, because in order for $f$ to be continuous at 0, it has to be bounded near 0, but your claim that $\lim_{x \rightarrow 0^+}f(x)=\infty$ means that $f$ is unbounded at 0, contradiction. I think you mean that $\lim_{x \rightarrow 1^-}f(x)=\infty$. Then, you get the limit as $\infty$ from your computation.

P.S. From one of your steps, $\sqrt{f'(x)^2} \neq f'(x)$, but it is $|f'(x)|$. So I think you just need to change that equality into inequality

Edit in response to edited question: Then, once we have $\int_0^1 \sqrt{f'(x)^2}=\int_0^1 |f'(x)| \geq |\lim_{t \rightarrow 0^+}\int_t^1 f'(x)|=|f(1)-\lim_{x \rightarrow 0^+}f(x)|=|-\infty|=\infty$

This corrects and finishes the proof.