# Thread: The Basics of Limits -- Calculus Intro

1. ## The Basics of Limits -- Calculus Intro

So the question is (and sorry for butchering this... I have tried Latex but cannot figure it out for the life of me)

limit[x:1-,-3x+2] (the limit on (-3x+2) as x approaches 1 from the left)
(PS. Tell me if this notation is acceptable, or my wording is bad...)

The book we use provides solutions, and the solutions show simply subbing in 1 to the (-3x+2), but that wouldn't work if it was piecewise, right?

So how would you do this other than using values like .999999 ?

2. you're finding a one-sided limit, so no problem ... do like the book says.

$\lim_{x \to 1^-} (-3x+2) = -1$

3. Okay, I understadn that as long as the domain is not restricted, that works out fine... But what if it was restricted so that x>1 ..... Then this wouldn't work out, correct? (And to tell you the WHOLE truth, that is what the book does... The question is actually a piecewise, and it just subs in 1 from both sides)

4. seems like some information is missing here ...

how is the piece-wise function defined in the first place, and what limit does the problem want you to find?

working with specifics is easier to explain.

5. I am sorry, I do not have the book on me anymore. I will likely check with the instructor tomorrow but... I'll see if I remember this all.

essentially it was:

f(x)= {-3x+2}, x<1 and {x^2}, x>=1

The questions asks us to find the limits from both directions, at x=1 ...

Then the SOLUTION shows them subbing 1 into the first side of the piecewise function .... That doesn't seem right to me.... 1 is not in the domain of that question so subbing it into there will not return the correct y value.

6. Originally Posted by mike_302
I am sorry, I do not have the book on me anymore. I will likely check with the instructor tomorrow but... I'll see if I remember this all.

essentially it was:

f(x)= {-3x+2}, x<1 and {x^2}, x>=1

The questions asks us to find the limits from both directions, at x=1 ...

Then the SOLUTION shows them subbing 1 into the first side of the piecewise function .... That doesn't seem right to me.... 1 is not in the domain of that question so subbing it into there will not return the correct y value.
ok, so here's the thing. with limits, we don't actually care about being "at" the point, but about being "close" to the point. now $\lim_{x \to 1^-} f(x)$ is asking, when $x$ gets "close" to 1 coming from the left (meaning, the x-values are always a little less than 1 but can get arbitrarily close to 1) what value does $f(x)$ get "close" to?

now, your piecewise function is defined as $-3x + 2$ for $x < 1$. so your left hand limit is concerned with this definition of $f(x)$. now, as long as the funciton is continuous, you can just plug in the value. so you plug in 1 into $-3x + 2$

Now, if they asked for $\lim_{x \to 1^+}f(x)$, they are asking for the limit as $x$ gets close to one from the right. hence you look at values where $x$ is a little bigger than 1. so you look for how $f(x)$ is defined for $x > 1$. here, we have $f(x) = x^2$ for $x \ge 1$, so that

$\lim_{x \to 1^+}f(x) = \lim_{x \to 1^+} (x^2) = 1^2 = 1$

7. also note ... if the left limit = the right limit, then the overall limit exists and has the value equal to the left and right limits.