Recent content by ppham27

Yes. Try \{1,x,x^3+x^2,x^3\}.

Try writing f(x) = 1 + [f(x) - 1]. Also, note that (1+u) = [(1+u)^{1/u}]^u.

Because \lim_{x \rightarrow 0}\frac{x}{\sin x} = \lim_{x \rightarrow 0}\frac{1}{\frac{\sin x}{x}} = \frac{\lim_{x \rightarrow 0}1}{\lim_{x \rightarrow 0}\frac{\sin x}{x}} = \frac{1}{1} = 1.

Sorry, I should clarify that it's true provided that limits of f and g exist at p and the limit of g at p is nonzero.

I'd also note that \lim_{x \rightarrow p} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow p} f(x)}{\lim_{x \rightarrow p} g(x)}.

As far as I can see, L'Hopital's rule doesn't even apply since the numerator does not equal 0 if you plug in 0. a^0 - a^(-0) - 2 = 1 - 1 - 2 = -2. You're simply dividing by a small positive number, so the limit is negative infinity.