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

Step 1
Consider the limit
${\displaystyle \underset{t\to 1}{lim}\frac{t^4-t-2}{t-1}}$
We want to represent it as a derivative of some function f at some number a, ${\displaystyle \ln }$ light of definition 5 we want to write our limit in the form
${\displaystyle \underset{t\to a}{lim}\frac{f\left(t\right)-f\left(a\right)}{t-a}}$
Since ${\displaystyle t-1}$ appears in the denominator, and ${\displaystyle t\to 1}$, we are motivated to take ${\displaystyle a=1}$
Looking at the numerator of the limit in question we want ${\displaystyle f\left(t\right)=t^4+t}$. Let us see if this is the right choice. By definition 5 we have
${\displaystyle f^{\prime }\left(1\right)=\underset{t\to 1}{lim}\frac{f\left(t\right)-f\left(1\right)}{t-1}}$
${\displaystyle =\underset{t\to 1}{lim}\frac{t^4+t-(1^4+1)}{t-1}}$
${\displaystyle =\underset{t\to 1}{lim}\frac{t^4+t-2}{t-1}}$
Which is what we wanted to establish.
Ans: ${\displaystyle f\left(t\right)=t^4+t,a=1}$

Step 1
Definition
${\displaystyle f^{\prime }\left(a\right)=\underset{x\to a}{lim}\frac{f\left(x\right)-f\left(a\right)}{x-a}}$
Step 2
${\displaystyle \underset{t\to 1}{lim}\frac{t^4+t-2}{t-1}}$
${\displaystyle \underset{t\to 1}{lim}\frac{(t^4+t)-(1^4+1)}{t-1}}$
${\displaystyle \underset{t\to 1}{lim}\frac{f\left(t\right)-f\left(1\right)}{t-1}=f^{\prime }\left(1\right)}$
Where ${\displaystyle f\left(t\right)=t^4+t}$
Note that ${\displaystyle a=1}$