How do I compute $\displaystyle\lim _{x\to 0} \tfrac{e^x+\sin x -1}{\ln(1+x)}$? I'm a Calculus I teacher's assistant. One of my students asked me how to compute this limit
$$\lim _{x\rightarrow 0} \dfrac{e^x+\sin x -1}{\ln(1+x)}$$
I could not solve it. I need some hint.
P.s: I'm not supposed to solve this using differentiation. My classes haven't entered that yet.
Thanks.
 A: Using Taylor series:
$$\lim_{x\to 0}\frac{e^x+\sin x -1}{\log(1+x)}=\lim_{x\to 0}\frac{(1+x)+x-1+o(x)}{x+o(x)}=2.$$
A: Use l'hospital's rule:
$$\lim _{x\rightarrow 0} \dfrac{e^x+\sin x -1}{\ln(1+x)}$$
$$\lim _{x\rightarrow 0} \dfrac{e^x+\cos x}{\frac{1}{1+x}}$$
$$\lim _{x\rightarrow 0} (e^x + \cos x)(x+1) = (1+1)*(1) = 2$$
Comment if you have questions.
A: Note that $\displaystyle\lim_{x \to 0}\dfrac{e^x+\sin x - 1}{\ln(1+x)} = \lim_{x \to 0}\dfrac{\dfrac{e^x-1}{x} + \dfrac{\sin x}{x}}{\dfrac{\ln(1+x)}{x}} = \dfrac{\left[\lim\limits_{x \to 0}\dfrac{e^x-1}{x}\right] + \left[\lim\limits_{x \to 0}\dfrac{\sin x}{x}\right]}{\left[\lim\limits_{x \to 0}\dfrac{\ln(1+x)}{x}\right]}$. 
Can you evaluate those 3 limits using only stuff that the students have seen? This will depend on how your book defines the functions $\sin x$, $e^x$, and $\ln x$. In most books $\displaystyle\lim_{x\to 0}\frac{\sin x}{x} = 1$ is shown geometrically (without the need for derivatives). I'm not sure how your book defines $e^x$ and $\ln x$.
