Q. 14 - Calculus

Question

What is the definition of the number e? What does this definition tell you about $\underset{h\to 0}{\lim }\frac{e^h-1}{h}$ ? Why is this limit relevant to calculating the derivative of the function $f\left(x\right)=e^x?$

Step-by-Step Solution

Answer

The number $e$ is also known as Euler's number is a mathematical constant approximately equal to $2.71828$ and  $\underset{h\to 0}{\lim }\frac{e^h-1}{h}=1$ and it is relevant to calculating the derivative of the function $f\left(x\right)=e^x.$

Step 1. Given information

We need to write the definition of the number $e,$ and also what does this tell us about $\underset{h\to 0}{\lim }\frac{e^h-1}{h}=1$ and why this limit is relevant to calculating the derivative of the function $f\left(x\right)=e^x.$

Step 2. Explanation

The number $e$ is also known as Euler's number is a mathematical constant approximately equal to $2.71828$ which can be characterized in many ways.

We know that,

$e=\underset{h\to 0}{\lim }{\left(1+h\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$h$}\right.}.$

For the sufficiently small value of $h$ we have.

$e\approx {\left(1+h\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$h$}\right..}$

$e\approx 1+h.$

$e^h-1\approx h.$

$\frac{e^h-1}{h}\approx 1.$

Since the preceding approximates get better as $h\to 0,$ it is the reason $\underset{h\to 0}{\lim }\frac{e^h-1}{h}=1.$

Thus,

$f\left(x\right)=e^x.$

$f^{\text{'}}\left(x\right)=\frac{d}{dx}\left(e^x\right).$

        $=\underset{h\to 0}{\lim }\frac{e^{x+h}-e^x}{h}.$

        $=\underset{h\to 0}{\lim }\left[e^x\left(\frac{e^h-1}{h}\right)\right].$

        $=e^x\left(\underset{h\to 0}{\lim }\frac{e^h-1}{h}\right).$

         $=e^x\times 1.$

         $=2.$

So, $f^{\text{'}}\left(x\right)=e^x.$