Question

Explain if the limit of the function f(t) exist at t=1 where,

F(t)={ t²+5 if t<1} 

 {3-3t if t>1}

Answer 1

The limit of $f(t)$ at $t=1$ exists if the one-sided limits exist at $t=1$ , and are equal.

For given function

$f(t)=\left\{\begin{array}{ll}t^2+5,&t<1,\\ 3-3t,&t>1\end{array}\right.$

the left-hand limit equals

$\displaystyle\lim\limits_{t\to1^-}f(t)= \lim\limits_{t\to1}(t^2+5)=1^2+5=1+5=6$,

the right-hand limit equals

$\displaystyle\lim\limits_{t\to1^+}f(t)= \lim\limits_{t\to1}(3-3t)=3-3\cdot1=3-3=0$.

So, the one-sided limits of given function exist at $t=1$ , but are unequal

$\displaystyle\lim\limits_{t\to1^-}f(t)\neq\lim\limits_{t\to1^+}f(t)$.

Hence, the limit of the function at $t=1$ does not exist.