Continuity of $g$ at $a=1$: Proven

[karush]

use the definition of continuity and the properties of limits to show that the function is continuous at the given number $a$
$$g(t)=\frac{t^2+5t}{2t+1}\qquad a=1$$
ok i assume we just plug in a for t
$$\frac{1^2+5(1)}{2(1)+1)}=\frac{6}{3}=2$$

theorem 4 if f and g are continuous at a and if c is a constant, then the following are functions are also continuous at a
\begin{align}\displaystyle
&f+g \quad f-g \quad cf \quad fg \quad\frac{f}{g}
\end{align}

 

[HOI]

The problem said "use the definition of continuity and the properties of limits" so you cannot just appeal to a theorem about continuity.

The "definition of continuity" is
"A function, f, is continuous at x= a if and only if
1) f(a) is defined
2) \lim_{x\to a} f(x) is defined
3) \lim_{x\to a} f(x)= f(a)
(Since (3) requires that the two sides of the equation be defined often only (3) is given.)

Here f(t)= \frac{t^2+ 5t}{2t+ 1} and a= 1. Yes, f(1)= \frac{1+ 5}{2+ 1}= \frac{6}{3}= 2 as you have shown so f(a) exists. Now what is \lim_{t\to a} f(t)= \lim_{t\to 1}\frac{t^2+ 5t}{2t+ 1}? You can use the "properties of limits" one of which is
"\lim_{x\to a} \frac{f(x)}{g(x)}= \frac{\lim_{x\to a} f(x)}{\lim_{x\to a} g(x)} provided both those limits exist and \lim_{x\to a} g(x)\ne 0."