Question

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

Answer 1

Hint: We have an expression whose value we have to find under the given limits and we have to solve this specifically using hospital’s rule. Firstly, in order to use the hospital’s rule, we have to make the value of the expression as either \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\]. Then, we will use the l’Hospital’s rule to get the value of the expression.

Complete step by step solution:
According to the given question, we have an expression whose value we have to find using l’Hospital’s rule. We have been given the limit as $t \to 0$.
L’Hospital’s rule states that for limits that gives indeterminate forms such as \[\dfrac{0}{0}\] or \[\dfrac{\infty }{\infty }\], we can solve the expression by using the formula:
\[\displaystyle \lim_{x\to c}\dfrac{f(x)}{g(x)}=\displaystyle \lim_{x\to c}\dfrac{f'(x)}{g'(x)}\]
We will start with our solution.
Let us suppose,
\[L=\displaystyle \lim_{t \to 0}\dfrac{{{e}^{t}}-1}{t}\]-----(1)
The limit given to us $t \to 0$, here \[t\] only approaches 0 and does not get the value of \[t=0\].
But we can see that, when \[t=0\] the expression gives us the indeterminate form that is, \[\dfrac{0}{0}\].
\[L=\dfrac{{{e}^{0}}-1}{0}=\dfrac{0}{0}\]
So, we can use l’Hospital’s rule for $t \to 0$.
We will be using the l’Hospital’s formula first and then we will be applying the limits.
 Using the formula, we get,
\[L=\displaystyle \lim_{t \to 0}\dfrac{{{e}^{t}}-1}{t}\]
\[\Rightarrow L=\displaystyle \lim_{t \to 0}\dfrac{\dfrac{d}{dt}({{e}^{t}}-1)}{\dfrac{d}{dt}(t)}\]
We know that, derivative of \[{{e}^{t}}\] remains \[{{e}^{t}}\] only. 1 is a constant so it will get differentiated to 0. And \[t\] being the variable with respect to which the expression is differentiated, we get 1.
So, we have,
\[\Rightarrow L=\displaystyle \lim_{t \to 0}\dfrac{\dfrac{d}{dt}({{e}^{t}})-\dfrac{d}{dt}(1)}{\dfrac{d}{dt}(t)}\]
\[\Rightarrow L=\displaystyle \lim_{t \to 0}\dfrac{{{e}^{t}}-0}{1}\]
Now, we will apply the limits, we get,
\[\Rightarrow L=\displaystyle \lim_{t \to 0}{{e}^{t}}\]
\[\Rightarrow L={{e}^{0}}=1\]
Therefore, the value of the expression is 1.

Note:
The value that we obtained after solving the expression using l’Hospital’s rule can be generalized and can be used straight away without having to do all the above calculation. We have,
\[\displaystyle \lim_{t \to 0}\dfrac{{{e}^{t}}-1}{t}=1\]
Or
\[\displaystyle \lim_{x\to 0}\dfrac{{{e}^{x}}-1}{x}=1\]
Also, we need to remember that, in \[\displaystyle \lim_{t \to 0}\]the value of \[t\] do not get equal to 0 in reality, but gets only close enough to 0.