Q. 3 - Calculus

Question

Explain graphically why it makes sense that a limit $\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ of the indeterminate form $\frac{0}{0}$ would be related to $\underset{x\to c}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$.

Step-by-Step Solution

Answer

Limits of the indeterminate form $\frac{0}{0}$ are related to the limit of the derivatives of its numerator and denominator functions.

Step 1. Given information

We have to prove graphically that a limit $\underset{x\to c}{\lim }\frac{f\left(x\right)}{g\left(x\right)}$ of the indeterminate form $\frac{0}{0}$ would be related to $\underset{x\to c}{\lim }\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$.

Step 2. Proof of the given question graphically.

Let us take an example

$\underset{x\to 0}{\lim }\frac{x^2}{2^x-1}$

The value of the limit is in the form of $\frac{0}{0}$.

The graphs of $f\left(x\right)=x^2$ and $g\left(x\right)=2^x-1$ are

Since we are interested in a limit as $x\to 0$, we should focus graph around $x=0$.

Near $x=0$, the graph of $f\left(x\right)=x^2$ looks a lot like its horizontal tangent line $y=0$, and the graph of $g\left(x\right)=2^x-1$ looks a lot like its tangent line $y=x$.

So the behaviour of the quotient $\frac{f\left(x\right)}{g\left(x\right)}=\frac{x^2}{2^x-1}$ as $x\to 0$ to be similar to the quotient of the corresponding tangent lines $\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$ at $x=0$.