Let $f(x) = \exp\left(-2^{1/3}\right) - \exp\left(-2^\frac{x}{3x-1}\right)$. I want to compute $$ \lim_{x\rightarrow +\infty} e^x f(x) $$
which according to the grapher is $+\infty$. It's been a while since I've computed limits, and the limit I'm getting is $-\infty$ instead: \begin{align*} \lim_{x\rightarrow +\infty} e^x f(x) &= \lim_{x\rightarrow +\infty}\exp\left(-2^{1/3}\right)e^x - \exp\left(x-2^\frac{x}{3x-1}\right) \\ &= \lim_{x\rightarrow +\infty} \exp\left(-2^{1/3}\right)e^x - \exp\left(x\,(1-\frac{1}{x}\,2^\frac{x}{3x-1})\right) \\ &= \lim_{x\rightarrow +\infty} \exp\left(-2^{1/3}\right)e^x - \exp(x(1-0)) \\ &= \lim_{x\rightarrow +\infty} e^x \left(e^{-2^{1/3}}-1\right) \\ = -\infty. \end{align*}