Question

Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).

\(\mathop {\lim }\limits_{t \to 0} \frac{{{e^{5t}} - 1}}{t}\),\(t = \pm 0.5,\, \pm 0.1,\, \pm 0.01,\, \pm 0.001,\, \pm 0.0001\).

Short answer

The value of the limit is \(\mathop {\lim }\limits_{t \to 0} \frac{{{e^{5t}} - 1}}{t} = 5\).

Step-by-step solution

Intuitive Definition of a Limit

Let \(f\left( x \right)\) be defined when \(x\) is near the number \(a\).(Which means that \(f\) is defined on some open interval that contains \(a\), except possibly at \(a\) itself.) Write it as:

\(\mathop {\lim }\limits_{x \to a} f\left( x \right) = L\)

The limit of \(f\left( x \right)\) as \(x\) approaches \(a\) is equal to \(L\). When the values of \(f\left( x \right)\) arbitrarily close to \(L\) by restricting \(x\) to be sufficiently close to \(a\) but not equal to \(a\).

Compare the definition 1 and 2 we obtain the following is true.

\(\mathop {\lim }\limits_{x \to a} f\left( x \right) = L\)if and only if \(\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = L\)and\(\mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = L\).

Evaluate the function at the given numbers

Let \(f\left( t \right) = \frac{{{e^{5t}} - 1}}{t}\).

Evaluate the function at \(x = 0.5\) as shown below:

\(\begin{aligned}\mathop {\lim }\limits_{t \to 0.5} \frac{{{e^{5t}} - 1}}{t} &= \frac{{{e^{5\left( {0.5} \right)}} - 1}}{{0.5}}\\ &= \frac{{12.182493 - 1}}{{0.5}}\\ &= \frac{{11.182493}}{{0.5}}\\ &= 22.364988\end{aligned}\)

The other values are obtained in a similar way listed in the table as shown below:

\(t\)	\(f\left( t \right)\)
\(\begin{aligned}0.5\\0.1\\0.01\\0.001\\0.0001\\ - 0.5\\ - 0.1\\ - 0.01\\ - 0.001\\ - 0.0001\end{aligned}\)	\(\begin{aligned}22.364988\\6.487213\\5.127110\\5.012521\\5.001250\\1.835830\\3.934693\\4.877058\\4.987521\\4.998750\end{aligned}\)

It is observed from the table thatleft, and right limits are the same and therefore, \(\mathop {\lim }\limits_{t \to 0} \frac{{{e^{5t}} - 1}}{t} = 5\).

Thus, the value of the limit is \(\mathop {\lim }\limits_{t \to 0} \frac{{{e^{5t}} - 1}}{t} = 5\).