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

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In summary: You have shown above that \lim_{t\to a} f(t) exists. What about \lim_{t\to a} g(t)? Again, g(1)= \frac{1^2+ 5(1)}{2(1)+ 1}= \frac{6}{3}= 2. Now, what about the limit as x goes to 1? Again, you have shown that f(1) exists but you have not shown that \lim_{t\to 1} g(t) exists. You can use the same properties of limits here. What is \lim_{t\to 1} g(t)= \lim_{t\to
  • #1
karush
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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}
 
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  • #2
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) [tex]\lim_{x\to a} f(x)[/tex] is defined
3) [tex]\lim_{x\to a} f(x)= f(a)[/tex]
(Since (3) requires that the two sides of the equation be defined often only (3) is given.)

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

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

What is continuity?

Continuity is a property of a function where there are no abrupt changes or breaks in the graph of the function. In other words, the function is smooth and connected at all points.

What does it mean for a function to be continuous at a specific point?

A function is continuous at a specific point if the limit of the function at that point exists and is equal to the value of the function at that point.

How do you prove continuity at a specific point?

To prove continuity at a specific point, we need to show that the limit of the function at that point exists and is equal to the value of the function at that point. This can be done using the epsilon-delta definition of continuity or by using the three-part definition of continuity.

What is the epsilon-delta definition of continuity?

The epsilon-delta definition of continuity states that for a function f to be continuous at a point a, for any given positive number epsilon, there exists a positive number delta such that the distance between the input x and the point a is less than delta, then the distance between the output f(x) and f(a) is less than epsilon.

What is the three-part definition of continuity?

The three-part definition of continuity states that a function f is continuous at a point a if the following three conditions are met: (1) f(a) is defined, (2) the limit of f(x) as x approaches a exists, and (3) the limit of f(x) as x approaches a is equal to f(a).

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