2 Q's: Limit and Continuity

[gaminlegend38]

Question 1:

if f(x)=x^2-2x, then f(2+h)=

I simply plugged the 2+h in for x, FOILed, and ended up getting h^2+2h, which is one of the multiple choice selections. However, it just seems too simple. Seems like I'm forgetting something here.

Question 2:

f(x)= { x^2-5x+2 if x>3
      { x+j      if x<=3

Find the value of j so that f(x) is continuous at x=3.

In this case, wouldn't j simply be zero? What I did was I plugged 3 in for x and solved for j (setting it equal to 3), which would obviously make j zero. Since the bottom would thus equal three, that would make it continuous, right? Once again, it seems too easy.

 

[o_O]

1. Nope, nothing wrong there.

2. Not quite. Look at it again. In order for f(x) to continuous at x = 3, you need:

\(\displaystyle \lim_{x \to 3} f(x) = f(3)\)

So, from the right hand side:

\(\displaystyle \lim_{x \to 3^{+}} f(x) \quad = \quad (3)^{2} - 5(3) + 2 \quad = \quad 9 - 15 + 2 \quad = \quad -4\)

That means the function f(3) = x + j must equal to -4 as this will make the original limit valid.

 

[gaminlegend38]

Okay, cool. Thanks for the help!