Find lim as x to 0 (ln(1+x)- (x²)/2+x)/(e^x- cos x)

Find $\displaystyle\lim_{x\to0}\frac{\ln(1+x)-\dfrac{x^{2}}{2}+x}{e^{x}-\cos x}$.

Step 1: Check the indeterminate form

At $x = 0$ the numerator is $\ln1-0+0 = 0$ and the denominator is $1-1 = 0$. The form is $\dfrac00$.

Step 2: Expand the numerator

From $\ln(1+x) = x-\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}-\cdots$, \[\ln(1+x)-\frac{x^{2}}{2}+x = \left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots\right)-\frac{x^{2}}{2}+x = 2x-x^{2}+\frac{x^{3}}{3}-\cdots \tag{1}\]

The two linear terms add, and the two quadratic terms add as well — hence the leading coefficient $2$.

Step 3: Expand the denominator

\[ e^{x}-\cos x = \left(1+x+\frac{x^{2}}{2}+\frac{x^{3}}{6}+\cdots\right)-\left(1-\frac{x^{2}}{2}+\frac{x^{4}}{24}-\cdots\right) = x+x^{2}+\frac{x^{3}}{6}+\cdots \tag{2} \]

The constants cancel, and the two halves of $x^{2}$ combine into a whole one.

Step 4: Form the ratio

From (1) and (2), for $x\neq0$ the common factor $x$ may be cancelled: \[\frac{2x-x^{2}+\dfrac{x^{3}}{3}-\cdots}{x+x^{2}+\dfrac{x^{3}}{6}+\cdots} = \frac{x\left(2-x+\dfrac{x^{2}}{3}-\cdots\right)}{x\left(1+x+\dfrac{x^{2}}{6}+\cdots\right)} = \frac{2-x+\dfrac{x^{2}}{3}-\cdots}{1+x+\dfrac{x^{2}}{6}+\cdots}.\]

Step 5: Evaluate the limit

Every remaining term carries a positive power of $x$, so letting $x\to0$: $\lim_{x\to0}\frac{2-x+\cdots}{1+x+\cdots} = \frac{2}{1} = 2$.

Step 6: Verification

\[\text{By l'H\^opital: } \lim_{x\to0}\frac{\dfrac{1}{1+x}-x+1}{e^{x}+\sin x} = \frac{1-0+1}{1+0} = 2 \;\;\checkmark\]

\[x = 0.01:\ 1.9702973; \qquad x = -0.01:\ 2.0303028; \qquad x = 0.001:\ 1.9970030 \;\;\checkmark\]

\[\text{Dividing the series: } \frac{2-x+\cdots}{1+x+\cdots} = 2-3x+3x^{2}-\cdots \;\;\checkmark\]