Question

evaluate the limit

Answer 1

\[\lim_{x \rightarrow 3^{-1}} \frac{ |4x-12| }{ x-3 }\]

Answer 2

You mean \(3^{-}\) without the negative one there, in the power?

Answer 3

\(\color{#000000}{ \displaystyle \lim_{x\to 3^{-}}\frac{|4x-12|}{x-3} }\)

Answer 4

Yep :)

Answer 5

I got 4 but that's not the right answer

I thought about taking numbers that are close to three since the function wouldn't be defined at 3

\[\frac{|4(3.1)-12| }{ 3.1-3 } = \frac{ 4/10 }{ 1/10 } = 4 \]

Answer 6

YOu are approaching 3 FROM THE LEFT

Answer 7

not from the right

Answer 8

That means all your values are little bit less than 3, although they do get closer and closer to 3.

Answer 9

oh I see~

Answer 10

And that means that means that 4(3) is always a little less than 12.
So, 4x-12 is always negative as \(\color{#000000}{ \displaystyle x\to 3^{-} }\).

So, that is an equivalent of:
\(\color{#000000}{ \displaystyle \lim_{x\to 3^{-}}\frac{-(4x-12)}{x-3} }\)

Answer 11

\[|4(2.9)-12|/(2.9-3) = -4 \]

Answer 12

\(\color{#000000}{ \displaystyle \lim_{x\to 3^{-}}\frac{-4(x-3)}{x-3} }\)

\(\color{#000000}{ \displaystyle \lim_{x\to 3^{-}}\frac{-4}{1}=-4 }\)

Answer 13

I got the convention coming from the left/right wrong.

Answer 14

Do you see what I did?
Just using algebra, you can evaluate the limit.

Answer 15

yep, you factored out a 4 which got you (x-3) and then the numerator and denominator canceled out.

Answer 16

what if you can't factor out anything how would you go about it

Answer 17

It all depends on a particular case.
There are some limits that don't exist, but, many of the once you are given would.

Answer 18

ones**

Answer 19

I don't have any examples other than that one. but I guess you would try to factor out when you can and apply limit laws

Answer 20

Yes.

Answer 21

Sometimes, you will be requiered to play with logarithms, and play for real....
They should get relevantly hard, but all limits, tho ar releva ntly straight-forward.

Answer 22

In any case,
Good Luck!

Answer 23

thank you for your help

Answer 24

np

Answer 25

Another Example (with some theory):

Consider the following function:
\(\color{#000000}{\displaystyle f(x)=x^4-2x^3+x^2+3x+2}\)
and suppose that you want to find
the equation of the tangent line to
this function at \(x=2\).

At first, since you don't know the
y-value of the point, let's find this
value first, to get the (full) point of
tangency. (Plug \(x=2\) into \(f(x)\).)

\(\color{#000000}{\displaystyle f(\color{blue}{2})=\color{blue}{2}^4-2\color{blue}{(2)}^3+\color{blue}{2}^2+3\color{blue}{(2)}+2}\)
\(\color{#000000}{\displaystyle f(2)=2^4-2^4+4+6+2}\)
\(\color{#000000}{\displaystyle f(2)=12}\)

------------------------------
■ Additional Calculation Note:
Note also that using basic algebra, I
can afford laziness. I just noted that
\(2(2)^3=2^1\cdot 2^3=2^{1+3}=2^4\). Now I
can cancel \(2^4\) without calculating it,
and that is exactly what I did.
\(\small \bf (\)The reason I'm making such a deal
from this, is that using similar tricks
makes calculations a lot easier.\(\bf\small )\)
------------------------------

Now, back to the problem.... You get,
\(\color{#000000}{\displaystyle f(2)=12}\), so the point of tangency is
\(\color{#000000}{\displaystyle (x,y)=(2,12)}\). (My notation of writing
points. Seems fancy and precise to me:))

Once you've found the point of tangency,
you need to find the slope of the tangent
line (which is the instantaneous slope of
the \(f(x)\) at \(x=2\)). Lastly, using the point
and the slope (when you find this slope),
find the equation of the tangent line.
(This last step is just algebra.)

------------------------------
Recall the definition of slope.
\(\color{#000000}{\displaystyle \frac{\Delta y}{\Delta x}=\frac{f(x+h)-f(x)}{h} }\)

(Shouldn't be hard to see how the numerator
represents the change in y (denoted by \(\Delta y\)),
and how denominator represents the change
in x (denoted by \(\Delta x\)).

Technically, and really, the slope is supposed
to be between two points. The question that
comes to mind then is:
So, if the slope is by definition between two
points, then how can there be such a thing as
«instantaneous slope» (or, slope at an instant)?

The answer is:
Actually, we are taking the slope between two points.
These points are: \(\color{#000000}{\displaystyle (x,f(x))}\) and \(\color{#000000}{\displaystyle (x+h,f(x+h))}\)
(Where \(\color{#000000}{\displaystyle h}\) is the \(\Delta x\), such that \(\color{#000000}{\displaystyle h\to 0 }\).)

The change in \(x\) (the \(h\)) is approaching zero. That is
- it is never actually going to be exactly equal zero,
but for any non-zero value that you choose to assign
to \(h\), (regardless of how close to zero this value,) the
\(h\) is going to be smaller (closer to zero).

\(\color{red}{*}\) The change in x (the \(h\)) is so small, this change in x
practically doesn't differ from a zero change, (or none
change) in x. Therefore, while we are technically finding
the slope between two points, which are: \(\color{#000000}{\displaystyle (x,f(x))}\) and
\(\color{#000000}{\displaystyle (x+h,f(x+h))}\), in practice we are really looking at a
slope at one point \(\color{#000000}{\displaystyle (x,f(x))}\) (since practically speaking,
\(\color{#000000}{\displaystyle (x,f(x))}\) and \(\color{#000000}{\displaystyle (x+h,f(x+h))}\) are really the same points.

For this reason we say that the
instantaneous slope at \(\color{#000000}{\displaystyle (x,f(x))}\) is.
\(\color{#000000}{\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} }\)

And that, expression is equivalent to the derivative f'(x).
\(\color{#000000}{\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} }\)
Another Example:

Consider the following function:
\(\color{#000000}{\displaystyle f(x)=x^4-2x^3+x^2+3x+2}\)
and suppose that you want to find
the equation of the tangent line to
this function at \(x=2\).

At first, since you don't know the
y-value of the point, let's find this
value first, to get the (full) point of
tangency. (Plug \(x=2\) into \(f(x)\).)

\(\color{#000000}{\displaystyle f(\color{blue}{2})=\color{blue}{2}^4-2\color{blue}{(2)}^3+\color{blue}{2}^2+3\color{blue}{(2)}+2}\)
\(\color{#000000}{\displaystyle f(2)=2^4-2^4+4+6+2}\)
\(\color{#000000}{\displaystyle f(2)=12}\)

------------------------------
■ Additional Calculation Note:
Note also that using basic algebra, I
can afford laziness. I just noted that
\(2(2)^3=2^1\cdot 2^3=2^{1+3}=2^4\). Now I
can cancel \(2^4\) without calculating it,
and that is exactly what I did.
\(\small \bf (\)The reason I'm making such a deal
from this, is that using similar tricks
makes calculations a lot easier.\(\bf\small )\)
------------------------------

Now, back to the problem.... You get,
\(\color{#000000}{\displaystyle f(2)=12}\), so the point of tangency is
\(\color{#000000}{\displaystyle (x,y)=(2,12)}\). (My notation of writing
points. Seems fancy and precise to me:))

Once you've found the point of tangency,
you need to find the slope of the tangent
line (which is the instantaneous slope of
the \(f(x)\) at \(x=2\)). Lastly, using the point
and the slope (when you find this slope),
find the equation of the tangent line.
(This last step is just algebra.)

------------------------------
Recall the definition of slope.
\(\color{#000000}{\displaystyle \frac{\Delta y}{\Delta x}=\frac{f(x+h)-f(x)}{h} }\)

(Shouldn't be hard to see how the numerator
represents the change in y (denoted by \(\Delta y\)),
and how denominator represents the change
in x (denoted by \(\Delta x\)).

Technically, and really, the slope is supposed
to be between two points. The question that
comes to mind then is:
So, if the slope is by definition between two
points, then how can there be such a thing as
«instantaneous slope» (or, slope at an instant)?

The answer is:
Actually, we are taking the slope between two points.
These points are: \(\color{#000000}{\displaystyle (x,f(x))}\) and \(\color{#000000}{\displaystyle (x+h,f(x+h))}\)
(Where \(\color{#000000}{\displaystyle h}\) is the \(\Delta x\), such that \(\color{#000000}{\displaystyle h\to 0 }\).)

The change in \(x\) (the \(h\)) is approaching zero. That is
- it is never actually going to be exactly equal zero,
but for any non-zero value that you choose to assign
to \(h\), (regardless of how close to zero this value,) the
\(h\) is going to be smaller (closer to zero).

\(\color{red}{*}\) The change in x (the \(h\)) is so small, this change in x
practically doesn't differ from a zero change, (or none
change) in x. Therefore, while we are technically finding
the slope between two points, which are: \(\color{#000000}{\displaystyle (x,f(x))}\) and
\(\color{#000000}{\displaystyle (x+h,f(x+h))}\), in practice we are really looking at a
slope at one point \(\color{#000000}{\displaystyle (x,f(x))}\) (since practically speaking,
\(\color{#000000}{\displaystyle (x,f(x))}\) and \(\color{#000000}{\displaystyle (x+h,f(x+h))}\) are really the same points.

For this reason we say that the
instantaneous slope at \(\color{#000000}{\displaystyle (x,f(x))}\) is.
\(\color{#000000}{\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} }\)

And that, expression is equivalent to the derivative f'(x).
\(\color{#000000}{\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h} }\)

So, the derivative is the instantaneous slope of the function.
Therefore, differentiate the function to find the (expression
for the) instantaneous slope, and then plug in \(x=2\) into the
derivative to find the instantaneous slope of \(f(x)\) at \(x=2\).

I'm going to differentiate the function, using the power rule,
and apply the rule that the derivative of constant is zero.
\(\color{#000000}{\displaystyle f(x)=x^4-2x^3+x^2+3x+2}\)
\(\color{#000000}{\displaystyle f'(x)=\color{blue}{4}x^{\color{blue}{4}-\color{red}{1}}-\color{blue}{3}\cdot 2x^{\color{blue}{3}-\color{red}{1}}+\color{blue}{2}x^{\color{blue}{2}-\color{red}{1}}+\color{blue}{1}\cdot 3x^{\color{blue}{1}-\color{red}{1}}+0}\)
\(\color{#000000}{\displaystyle f'(x)=4x^3-6x^2+2x+3}\)

Notice that this is the reason why derivative of ax (with
respect to x) is just a. (In this case, d/dx (3x) = 3)
And that is the expression that gives the slope for \(f(x)\),
at any point \(x\). So again, to find slope \(x=2\), plug in \(x=2\)
in \(f'(x)\).

\(\color{#000000}{\displaystyle f'(x)=4x^3-6x^2+2x+3}\)
\(\color{#000000}{\displaystyle f'(\color{red}{2})=4\color{red}{(2)}^3-6\color{red}{(2)}^2+2\color{red}{(2)}+3}\)
\(\color{#000000}{\displaystyle f'(2)=4(2)^3-3(2)^3 +7=1\cdot (2)^3+9=8+7=15}\)

Well, you didn't have to do it this way, I just considered
this calculation to show how certain tricks (like factoring)
can be applied. In any case, we have found that \(\color{#000000}{\displaystyle f'(2)=15}\).
(In other words, the instantaneous slope at x=2 is 15.)
------------------------------
We know the slope is \(15\), and the point is \((2,12)\), so
the line (using the point slope form) is:

\(\color{#000000}{\displaystyle y-y_o=f'(x_o)(x-x_o)\quad \Longrightarrow \quad y-12=15(x-2)}\)

If case the above formula confuses you, this is exactly
the same thing as \(\color{#000000}{\displaystyle y-y_1=m(x-x_1)}\), with the only
difference that I denoted the point using \((x_o,y_o)\),
instead of \(x_1,y_1\), and wrote my slope \(m\) as \(f'(x_o)\),
since the slope of a function is given by the derivative
of the function evaluated at a certain point.

You can leave the equation of the tangent line as it is,
or re-write it. That's entirely up to your taste!
\(\color{#000000}{\displaystyle y-12=15(x-2)}\)
\(\color{#000000}{\displaystyle y-12=15x-30}\)
\(\color{#000000}{\displaystyle y=15x-18}\)

Slope intercept form is just fine, and probably better.
(Not that it matters anyway, once you find any
version of the equation of the tangent line).
------------------------------
\(\color{red}{\bf In~Summary,~the~work~is:}\)

I want to find the tangent line to \(f(x)\) at \(x=2\).
\(\color{#000000}{\displaystyle f(x)=x^4-2x^3+x^2+3x+2}\)

The point:
\(\color{#000000}{\displaystyle f(x)=x^4-2x^3+x^2+3x+2}\)
\(\color{#000000}{\displaystyle f(2)=(2)^4-2(2)^3+(2)^2+3(2)+3}\)
\(\color{#000000}{\displaystyle f(2)=12}\)

The slope:
\(\color{#000000}{\displaystyle f(x)=x^4-2x^3+x^2+3x+2}\)
\(\color{#000000}{\displaystyle f'(x)=4x^3-6x^2+2x+3+0}\)
\(\color{#000000}{\displaystyle f'(2)=4(2)^3-6(2)^2+2(2)+3}\)
\(\color{#000000}{\displaystyle f'(2)=15}\)

The equation of the tangent line:
\(\color{#000000}{\displaystyle y-y_o=f'(x_o)(x-x_o)}\)
\(\color{#000000}{\displaystyle y-12=15(x-2)}\)
or, \(\color{#000000}{\displaystyle y=15x-18}\)
------------------------------

So, the derivative is the instantaneous slope of the function.
Therefore, differentiate the function to find the (expression
for the) instantaneous slope of \(f(x)\), and then plug \(x=2\)
into the derivative to find instantaneous slope of \(f(x)\) at \(x=2\).

I'm going to differentiate the function, using the power rule,
and apply the rule that the derivative of constant is zero.
\(\color{#000000}{\displaystyle f(x)=x^4-2x^3+x^2+3x+2}\)
\(\color{#000000}{\displaystyle f'(x)=\color{blue}{4}x^{\color{blue}{4}-\color{red}{1}}-\color{blue}{3}\cdot 2x^{\color{blue}{3}-\color{red}{1}}+\color{blue}{2}x^{\color{blue}{2}-\color{red}{1}}+\color{blue}{1}\cdot 3x^{\color{blue}{1}-\color{red}{1}}+0}\)
\(\color{#000000}{\displaystyle f'(x)=4x^3-6x^2+2x+3}\)

Notice that this is the reason why derivative of ax (with
respect to x) is just a. (In this case, d/dx (3x) = 3)
And that is the expression that gives the slope for \(f(x)\),
at any point \(x\). So again, to find the instantaneous slope
at \(x=2\), plug \(x=2\) into \(f'(x)\).

\(\color{#000000}{\displaystyle f'(x)=4x^3-6x^2+2x+3}\)
\(\color{#000000}{\displaystyle f'(\color{red}{2})=4\color{red}{(2)}^3-6\color{red}{(2)}^2+2\color{red}{(2)}+3}\)
\(\color{#000000}{\displaystyle f'(2)=4(2)^3-3(2)^3 +7=1\cdot (2)^3+9=8+7=15}\)

Well, you didn't have to do it this way, I just considered
this calculation to show how certain tricks (like factoring)
can be applied. In any case, we have found that \(\color{#000000}{\displaystyle f'(2)=15}\).
(In other words, the instantaneous slope at x=2 is 15.)
------------------------------
We know the slope is \(15\), and the point is \((2,12)\), so
the line (using the point slope form) is:

\(\color{#000000}{\displaystyle y-y_o=f'(x_o)(x-x_o)\quad \Longrightarrow \quad y-12=15(x-2)}\)

If case the above formula confuses you, this is exactly
the same thing as \(\color{#000000}{\displaystyle y-y_1=m(x-x_1)}\), with the only
difference that I denoted the point using \((x_o,y_o)\),
instead of \(x_1,y_1\), and wrote my slope \(m\) as \(f'(x_o)\),
since the slope of a function is given by the derivative
of the function evaluated at a certain point.

You can leave the equation of the tangent line as it is,
or re-write it. That's entirely up to your taste!
\(\color{#000000}{\displaystyle y-12=15(x-2)}\)
\(\color{#000000}{\displaystyle y-12=15x-30}\)
\(\color{#000000}{\displaystyle y=15x-18}\)

Slope intercept form is just fine, and probably better.
(Not that it matters anyway, once you find any
version of the equation of the tangent line).
------------------------------
\(\color{red}{\bf In~Summary,~the~work~is:}\)

I want to find the tangent line to \(f(x)\) at \(x=2\).
\(\color{#000000}{\displaystyle f(x)=x^4-2x^3+x^2+3x+2}\)

The point:
\(\color{#000000}{\displaystyle f(x)=x^4-2x^3+x^2+3x+2}\)
\(\color{#000000}{\displaystyle f(2)=(2)^4-2(2)^3+(2)^2+3(2)+3}\)
\(\color{#000000}{\displaystyle f(2)=12}\)

The slope:
\(\color{#000000}{\displaystyle f(x)=x^4-2x^3+x^2+3x+2}\)
\(\color{#000000}{\displaystyle f'(x)=4x^3-6x^2+2x+3+0}\)
\(\color{#000000}{\displaystyle f'(2)=4(2)^3-6(2)^2+2(2)+3}\)
\(\color{#000000}{\displaystyle f'(2)=15}\)

The equation of the tangent line:
\(\color{#000000}{\displaystyle y-y_o=f'(x_o)(x-x_o)}\)
\(\color{#000000}{\displaystyle y-12=15(x-2)}\)
or, \(\color{#000000}{\displaystyle y=15x-18}\)
------------------------------
The end. (Apologize for a long response :))