Find lim as t to 0 [frac1t- (ln(1+t))/(t²) ]

Find $\displaystyle\lim_{t\to0}\left[\frac1t-\frac{\ln(1+t)}{t^{2}}\right]$.

Step 1: Identify the indeterminate form

As $t\to0$ the term $\dfrac1t$ becomes infinite, and so does $\dfrac{\ln(1+t)}{t^{2}}$, since $\ln(1+t)\sim t$ makes it behave like $\dfrac1t$. Both carry the same sign on each side of the origin, so the difference is of the form $\infty-\infty$.

Step 2: Combine into a single fraction

\[ \frac1t-\frac{\ln(1+t)}{t^{2}} = \frac{t-\ln(1+t)}{t^{2}}. \tag{1} \]

At $t = 0$ the numerator is $0-\ln1 = 0$ and the denominator is $0$: the form is $\dfrac00$.

Step 3: Expand using the Maclaurin series

\[ \ln(1+t) = t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\cdots \;\Longrightarrow\; t-\ln(1+t) = \frac{t^{2}}{2}-\frac{t^{3}}{3}+\cdots \tag{2} \]

The leading term of the numerator is of order $t^{2}$ — exactly the order of the denominator.

Step 4: Form the ratio

\[ \frac{t-\ln(1+t)}{t^{2}} = \frac{\dfrac{t^{2}}{2}-\dfrac{t^{3}}{3}+\cdots}{t^{2}} = \frac12-\frac{t}{3}+\cdots \]

Step 5: Evaluate the limit

\[ \lim_{t\to0}\left(\frac12-\frac{t}{3}+\cdots\right) = \frac12. \]

Step 6: Verification

\[t = 0.1:\ 0.4689820; \qquad t = -0.1:\ 0.5360516; \qquad t = 0.01:\ 0.4966915 \;\;\checkmark\]

\[\text{Series } \frac12-\frac{t}{3}+\frac{t^{2}}{4}:\ \text{at } t = 0.1,\ 0.5-0.03333+0.0025 = 0.46917 \;\;\checkmark\]

\[\text{Twice by l'H\^opital: } \frac{1-\frac{1}{1+t}}{2t} = \frac{1}{2(1+t)} \to \frac12 \;\;\checkmark\]