Question

How do you find $\lim \dfrac{\ln t}{t-1}$ as $t\to 0$ using l’Hospital’s Rule?

Answer 1

Hint: We have to use the l’Hospital’s Rule to find the solution to the given limits. Two different functions are given in the numerator and the denominator, thus we shall differentiate them separately. Then, we will apply the given value of the limit in the function of the numerator as well as the denominator and assemble them together to find our final solution.

Complete step-by-step solution:
In L’Hospital’s Rule, we use derivatives to find the limits that are in indeterminate form. The indeterminate, undefined forms are those in which the limit turns out to be of the form $\dfrac{0}{0}$, $\dfrac{\infty }{\infty }$ or $\dfrac{-\infty }{\infty }$ when we put the values of the limit.
According to l’Hospital’s Rule, if the limit $\displaystyle \lim_{x \to c}f\left( x \right)=0$, $\displaystyle \lim_{x \to c}g\left( x \right)=0$ and if $\displaystyle \lim_{x \to c}\dfrac{{f}'\left( x \right)}{{g}'\left( x \right)}=L$ exists where, L is any real number, then we can say that $\displaystyle \lim_{x \to c}\dfrac{f\left( x \right)}{g\left( x \right)}=L$.
Also, if the limit $\displaystyle \lim_{x \to c}f\left( x \right)=\pm \infty $, $\displaystyle \lim_{x \to c}g\left( x \right)=\pm \infty $ and if $\displaystyle \lim_{x \to c}\dfrac{{f}'\left( x \right)}{{g}'\left( x \right)}=L$ exists where, L is any real number, then we can say that $\displaystyle \lim_{x \to c}\dfrac{f\left( x \right)}{g\left( x \right)}=L$.
Given that $\lim \dfrac{\ln t}{t-1}$ as $t\to 0$.
Let $f\left( t \right)=\ln t$ and $g\left( t \right)=t-1$.
We shall first differentiate $f\left( t \right)=\ln t$ with respect to t.
$\Rightarrow \dfrac{d}{dt}f\left( t \right)=\dfrac{d}{dt}\ln t$
From basic differentiation, we know that $\dfrac{d}{dx}\ln x=\dfrac{1}{x}$. Applying this, we get
$\Rightarrow {f}'\left( t \right)=\dfrac{1}{t}$ ……………. Equation (1)
We shall now differentiate $g\left( t \right)=t-1$ with respect to t.
$\Rightarrow \dfrac{d}{dt}g\left( t \right)=\dfrac{d}{dt}\left( t-1 \right)$
From basic differentiation, we know that $\dfrac{d}{dx}x=1$. Applying this, we get
$\Rightarrow {g}'\left( t \right)=1$ ……………. Equation (2)
Combining (1) and (2), we get
$\Rightarrow \displaystyle \lim_{t\to 0}\dfrac{\ln t}{t-1}=\dfrac{\dfrac{1}{t}}{1}$
Applying the values of limit, we get
$\Rightarrow \displaystyle \lim_{t\to 0}\dfrac{\ln t}{t-1}=\dfrac{\dfrac{1}{0}}{1}$
Since, $\dfrac{1}{0}=\infty $, we get
$\Rightarrow \displaystyle \lim_{t\to 0}\dfrac{\ln t}{t-1}=\dfrac{\infty }{1}$
$\Rightarrow \displaystyle \lim_{t\to 0}\dfrac{\ln t}{t-1}=\infty $
Therefore, $\lim \dfrac{\ln t}{t-1}$ as $t\to 0$ using l’Hospital’s Rule is equal to $\infty $ or is not defined.

Note: Limits are the most basic part of calculus. We use limits to figure out derivatives of functions. In fact, the definition of a derivative uses the notion of a limit. It is a slope around the point as we take the limit of points closer and closer to the point in the question given.