Find the limit of (1 - cos x)/(1 - e^x) as x approaches 0.

Step-by-step solution

1. 1

   Confirm the form is $\tfrac{0}{0}$. At $x=0$ both parts vanish: $1-\cos 0=0$ and $1-e^{0}=0$. So the quotient is indeterminate and needs work — a common wrong instinct is to answer $1$ because "both go to zero at the same time", but they do not: they vanish at different rates.

2. 2

   Compare the rates with Maclaurin series. Keeping the leading term of each:

   $1-\cos x=\frac{x^2}{2}-\frac{x^4}{24}+\cdots,\qquad 1-e^{x}=-x-\frac{x^2}{2}-\cdots$

   The numerator is second order in $x$, the denominator only first order. That mismatch already tells you the limit is $0$.

3. 3

   Divide the leading behaviours.

   $\frac{1-\cos x}{1-e^{x}}=\frac{\frac{x^2}{2}\left(1-\frac{x^2}{12}+\cdots\right)}{-x\left(1+\frac{x}{2}+\cdots\right)}=-\frac{x}{2}\cdot\frac{1+O\!\left(x^2\right)}{1+O(x)}\longrightarrow 0$

   The bracketed factors both tend to $1$, so the whole thing behaves like $-\tfrac{x}{2}$ near $0$.

4. 4

   Verify with l’Hôpital’s rule. Differentiating numerator and denominator once:

   $\lim_{x\to 0}\frac{\sin x}{-e^{x}}=\frac{0}{-1}=0$

   One application is enough here, because after differentiating the form is $\tfrac{0}{-1}$, which is determinate. Applying the rule a second time would be a mistake — the rule only applies while the form is still indeterminate.

5. 5

   Check the two-sided behaviour numerically. At $x=10^{-3}$ the quotient is $-0.00049975$; at $x=10^{-5}$ it is $-5.00\times 10^{-6}$; at $x=-10^{-5}$ it is $+5.00\times 10^{-6}$. The values approach $0$ from opposite sides, consistent with the linear approximation $-\tfrac{x}{2}$, and the two-sided limit is

   $\lim_{x\to 0}\frac{1-\cos x}{1-e^{x}}=0$