How to I get the limit without using L’Hospital’s rule?

Question

Sorry I haven't figured out how to properly write up questions on this yet. I have to get the limit as $x$ goes to $1$, of the function $- (x-1)/\log(x)$ without using L’Hospital’s rule, any suggestions?

Answer 1

Setting $x-1=t$, you have to find the limit of $-\dfrac{t}{\ln(1+t)}$ as $t$ tends to $0$, and a well-known high school result asserts that $$\lim_{t\to 0}\frac{\ln(1+t)}t=1,$$ i.e. the slope of the curve $y=\ln x$ at the point $(1,0)$ is equal to $1$.

Answer 2

You use the definition of the derivative:

$$ \left.\frac{d}{dt}\log(t)\right|_{t=1} = \lim_{x \to 1} \frac{\log(x) - \log(1)}{x-1}. \tag{$*$} $$

From there, how you compute the derivative depends on how you define $\log$. Two common definitions are:

1. $\log(x) = \displaystyle \int_{1}^x \frac1t \;dt$. Then you compute $(*)$ using the Fundamental Theorem of Calculus.
2. $\log(x)$ is the inverse function to $e^x$. Then you compute $(*)$ using the formula $$ \frac{d}{dt}f^{-1}(y) = \frac{1}{f'(x)}\, \text{ where } y = f(x). $$ In this case, the derivative is $1/e^0 = 1$.

Answer 3

Substitute $x=e^y$, so the limit becomes $$ \lim_{y\to0}-\frac{e^y-1}{y}=-1 $$ If you aren't allowed to use l’Hôpital, you surely are allowed to use some “well-known” limits, aren't you?