 # Each limit represents the derivative of some function f at some number a. 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.

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,

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