Question

The line defined by the equation 2y+3=-2/3(x-3) is tangent to the graph of g(x) at x=-3. What is the value of the limit as x approaches -3 (g(x)-g(-3))/(x+3)

Answer 1

2y+3=-2/3(x-3)
minus 3 both sides
2y=-2/3(x-3)-3
divide both sides by 2
y=-1/3(x-3)-3/2
ok, so one thing we can do is evaluate numbers super close to it
when x=3.00001, then the result is aprox -1.5
when x=2.99999, the result is -1.5
the value of the limit is -1.5 or -3/2

Answer 2

The limit has a value of -1/3

Explanation:

The derivative of a curve at a point represents the slope of the tangent line at a point. Thus if we can obtain the slope of the given line equation that is tangent to the graph g(x), we will know the value of the limit.

Since the limit of the difference quotient represents the derivative which is the slope of the tangent line.

Identifying the slope of the tangent line

We can proceed solving for y from the given equation

`2y+3= -\cfrac 23 (x-3)`

Subtracting by 3 both sides and distributing -2/3 to each term inside the parenthesis give us

`2y= -\cfrac 23 x+2-3`

`2y= -\cfrac 23 x-1`

Then we can divide both sides by 2

`y= -\cfrac 13 x-\cfrac12`

We have the equation in slope-intercept form, y = mx+b, so we can tell that the slope is -1/3, so we can write

`\displaystyle \lim_(x\to -3)\cfrac{g(x)-g(-3)}{x+3}= -\cfrac 13`

Thus the limit of the difference quotient is -1/3 as well since it represents the slope of the tangent line.