Find the number $a$ and the function $f$ given the following limit:

\[\lim_{t\to 1} \frac{t^4 + t – 2}{t-1}\]

The aim of this question is to learn the **differentiation** (calculation of derivative) from **first principles** (also called by definition or by **ab-initio method**).

To solve this question, one needs to know the **basic definition of a derivative**. The derivative of a function $f(x)$ with respect to an independent variable $x$ is the defined as a function $f′(x)$ described by following equations:

**Equation 1: **Most fundamental definition

\[f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}\]

**Equation 2: **Same value can be calculated by using any number $a$ through following limit formula:

\[f'(x) = \lim_{x\to a} \frac{f(x)-f(a)}{x – a}\]

To solve such questions, we simply need to **convert/rearrange the given limit function** into such a form that it matches any one of the above equations. Once we have a similar-looking equation, we can find the values of the number $a$ and the function $f$ by a simple comparison.

It can be noted that both definitions or equations represent the same concept so one can see the denominator of the given limit function and the limit value to guess which equation is most suitable. For example, if there is only one number in the denominator and the **limit approaches zero, we use equation no. 1**. However, we may **consider equation no. 2 if the limit approaches a number** or there is a variable term in the denominator.

## Expert Answer

The equation given in the question represents some **derivative** $f'(t)$.

\[f'(t) = \lim_{t\to 1} \frac{t^4 + t – 2}{t-1}\]

Let’s just **re-arrange**/manipulate the given **limit** for achieving this purpose,

\[f'(t) = \lim_{t\to 1} \frac{t^4 + t – (2)}{t-1}\]

\[f'(t) = \lim_{t\to 1} \frac{t^4 + t – (1+1)}{t-1}\]

\[f'(t) = \lim_{t\to 1} \frac{t^4 + t – (1^4 + 1)}{t-1}\]

Now, if we **replace $a = 1$** in above equation,

\[f'(t) = \lim_{t\to a} \frac{t^4 + t – (a^4 + a)}{t-a}\]

Which looks **very similar to the 2nd equation** of the definition of the derivative.

## Numerical Result

So the solution to the given **equation** is:

\[f(x) = x^4-x \text{ with } a = 1\]

## Example

If the following **limit** represents the **derivative** of some **function** $f$ at some number $a$. Find the number $a$ and the **function** $f$.

\[\lim_{h\to 0} \frac{\sqrt{9+h}-3}{h}\]

The equation given in the question represents some **derivative** $f'(x)$.

\[f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}\]

**Rearranging** the limit:

\[f'(x) = \lim_{h\to 0} \frac{\sqrt{9+h}-3}{h} \]

\[f'(x) = \lim_{h\to 0} \frac{\sqrt{9+h}-\sqrt{9}}{h}\]

\[f'(x) = \lim_{h\to 0} \frac{f(9+h)-f(9)}{h}\]

Now, if we **replace $x = 9$** in above equation:

\[f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}\]

Which looks very **similar to the 1st equation** of the definition of the **derivative.** So,

\[f(x) = \sqrt{x} \text{ with } a = 9\]