The question is to prove the following limit, $n \to \infty $
$$ \frac{n}{t} \ln({1 + \frac{t}{n}}) \to 1 $$
I could do this using something like L'Hôpital's rule but the hint is "Use the differentiability of log at 1 to show that for each t"
I don't know how to prove it using the hint. Any help would be appreciated?
On
As an alternative way, note that as $n \to \infty,$ we have $r:=n/t\to \infty,$ so the limit is the same as
$$\lim_{r \to\infty }r\ln (1+1/r)=\ln \lim_{r \to\infty }(1+1/r)^r=\ln e =1, $$
where the first equality is by limit and log properties, and the second equality is by the limit definition of $e$.
Note that the limit in question is equivalent to $$ \lim\limits_{n\to\infty} \frac{\ln\left(1+\frac{t}{n}\right)}{\frac{t}{n}}. $$ Suppose we fixed a value of $t$, then we note that $\frac{t}{n}\to 0$ as $n\to\infty$. By a change of variables $h = \frac{t}{n}$, we rewrite the limit as $$ \lim\limits_{h\to 0} \frac{\ln\left(1+h\right)}{h}. $$ Now use the hint. (Can you think of this limit as the derivative of something somewhere?) Note that this limit works for every fixed value of $t$, so this should work for every $t$, and whenever $n\to\infty$.
Edit:
The expression above is exactly the derivative of the natural log function at $x = 1$. Indeed, by definition, the derivative is given by \begin{align*} [\ln(x)]'= \lim\limits_{h\to 0} \frac{\ln (x+h)-\ln(x)}{(x+h)-x}, \end{align*} then when evaluated at $x = 1$, we have \begin{align*} [\ln(x)]'\mid_{x = 1}&= \lim\limits_{h\to 0} \frac{\ln (1+h)-\ln(1)}{(1+h)-1}\\ &= \lim\limits_{h\to 0} \frac{\ln (1+h)}{h} \end{align*} Also note that the value of this limit should be well-known...