# limits and continuity

• Feb 19th 2009, 05:17 PM
Yamahammer342
Let F be a function defined by f(x)=2x-3 x≤4
1+x x>4

A. Find the limit of F(x) as x approaches 4 from the right

B. Is F continuous on the interval from x=0 to x=10? why or why not

Is there any way to do this without a graphing calc? if there is i cant figure it out.

This didnt really come out the way i wanted F(x) is suppost to be defined by both of these 2x-3 x≤4 and 1+x x>4. i just cant type it like it is on the paper
• Feb 19th 2009, 08:29 PM
mollymcf2009
Quote:

Originally Posted by Yamahammer342
Let F be a function defined by f(x)=2x-3 x≤4
1+x x>4

A. Find the limit of F(x) as x approaches 4 from the right

B. Is F continuous on the interval from x=0 to x=10? why or why not

Is there any way to do this without a graphing calc? if there is i cant figure it out.

f(x) = { $\displaystyle ^{2x-3}_{1+x}$ if $\displaystyle ^{x \leq 4}_{x>4}$

A. $\displaystyle \lim_{x \rightarrow 4^+} 1 + x = 5$

Because as $\displaystyle x \rightarrow 4^+$ concerns positive numbers greater than 4, you can use the 1+x part of the function to find the limit without having to graph it

B. This is a piecewise function. Based on the conditions of the function, a piecewise function can be continuous or discontinuous depending on whether the stipulations given cause there to be a discontinuity. In the case of this piecewise function, it IS continuous on the interval (0,10) because there is no break in the graph. The two pieces of this function intersect at x=4. Should the function have read as:

f(x) = { $\displaystyle ^{2x-3}_{1+x}$ if $\displaystyle ^{x<4}_{x>4}$
Then the function would not be continuous, because x = 2 would not be included in the domain of the function. There would be a hole at x=2
• Feb 20th 2009, 02:51 AM
mr fantastic
Quote:

Originally Posted by mollymcf2009
[snip]
Should the function have read as:

f(x) = { $\displaystyle ^{2x-3}_{1+x}$ if $\displaystyle ^{x<4}_{x>4}$
Then the function would not be continuous, because x = 2 would not be included in the domain of the function. There would be a hole at x=2

Actually, the above function is continuous at x = 2. There's no hole at x = 2.
• Feb 20th 2009, 01:46 PM
mollymcf2009
Quote:

Originally Posted by mr fantastic
Actually, the above function is continuous at x = 2. There's no hole at x = 2.

I meant at 4, not 2. Sorry!