I have to compute the limit of $ f(x)=\frac{2+x+e^x}{2+x-e^x}$ when $x$ approaches $\pm \infty$. My question is about the leading term. Am I right if I consider that $e^x$ is my leading term when $x \to +\infty$ and then I could factor by this term to remove the IF like this ? $$\lim_{x\to +\infty} f(x)= \lim_{x\to +\infty}=\frac{e^x(\frac{2}{e^x}+\frac{x}{e^x}+1)}{e^x(\frac{2}{e^x}+\frac{x}{e^x}-1)} =-1$$
On the other hand, what about when $x\to -\infty$ ? Am I right if I consider in this case that $x$ is my leading term when $x \to -\infty$ ?
$$\lim_{x\to -\infty} f(x)= \lim_{x\to -\infty}=\frac{x(\frac{2}{x}+1+\frac{e^x}{x})}{x(\frac{2}{x}+1-\frac{e^x}{x})} = 1$$
EDIT : Is there another method to calculate the limit of this function when $x \to - \infty$ ?