Find lim as x to 0 (1- cos x)/(1- cosh x)
Find $\displaystyle\lim_{x\to0}\frac{1-\cos x}{1-\cosh x}$.
Explanation of Method
The circular and hyperbolic cosines differ only by a sign.
Step 1: The form
Both $\cos0$ and $\cosh0$ equal $1$, so numerator and denominator both vanish: the form is $\dfrac00$.
Step 2: Expand both
\[ \cos x = 1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\cdots \;\Longrightarrow\; 1-\cos x = \frac{x^{2}}{2}-\frac{x^{4}}{24}+\cdots, \]
\[ \cosh x = 1+\frac{x^{2}}{2}+\frac{x^{4}}{24}+\cdots \;\Longrightarrow\; 1-\cosh x = -\frac{x^{2}}{2}-\frac{x^{4}}{24}-\cdots \]
The two series are identical except that every term of the second carries a minus sign.
Step 3: Divide
\[ \frac{1-\cos x}{1-\cosh x} = \frac{x^{2}\left(\dfrac12-\dfrac{x^{2}}{24}+\cdots\right)}{x^{2}\left(-\dfrac12-\dfrac{x^{2}}{24}-\cdots\right)} \longrightarrow \frac{1/2}{-1/2} = -1. \]
Step 4: Verification
\[\text{Twice by l'H\^opital: } \frac{\sin x}{-\sinh x} \to \frac{\cos x}{-\cosh x} \to \frac{1}{-1} = -1 \;\;\checkmark\]
\[x = 0.1:\ \frac{0.0049958}{-0.0050042} = -0.998337 \;\;\checkmark\]
\[\text{Exact identity: } \frac{1-\cos x}{1-\cosh x} = -\frac{\sin^{2}(x/2)}{\sinh^{2}(x/2)} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\lim_{x\to0}\frac{1-\cos x}{1-\cosh x} = -1\,}\]
Remark
Remark. The last line makes the answer transparent: the half-angle identities give $1-\cos x = 2\sin^{2}\tfrac{x}{2}$ and $\cosh x-1 = 2\sinh^{2}\tfrac{x}{2}$, and since $\sin t/\sinh t\to1$ the square of that ratio tends to $1$ — leaving only the minus sign that the reversed subtraction introduced.