How to approach problems on limits

[helpm3pl3ase]

lim f(5 + h) − f(5)/h
h→0

given that f(5) = 4 and f  (5) = −95.

Iam not really sure how to approach this problem. I know its in 0/0 form so you can use L'hosp rules.


For another question Iam asked to find the equation for the tangent line of

xy 3 + y − 23 = x

where y = 2.

is this a related rates problem?? Did I approach this right:

(x)(3y^2 dy/dx) + (y^3)(1) + 1 dy/dx = 1
= 3y^2 dy/dx + 1 dy/dx = 1 - y^3
= (dy/dx) 3y^2x + 1) = 1 - y^3
dy/dx = (1-y^3)/(3y^2x + 1)

then x = 3 from original equation.
dy/dx = -7/12(3)+1
= -7/37

then

Tangent line = y = mx + b
y = -7/37(3) + 2??

Thanks for your help

 

[helpm3pl3ase]

anyone?

 

[Gib Z]

Calm down, calm down..ill help...

We'll I am working on the assumption that you f space (5) meant gradient at x=5?
Ok, We'll if you noticed, what your equation is describing is Differentiation by First Principles. In this case, rather than x, it used a specific case, x=5.
Since we know the gradient is -95 already, that is the solution to the equation.

For the 2nd one, looking at your working I'm thinking that's a y^3, not y space 3...well you just take all terms to one side, you get x(y^3-1) + (y-23) = 0. We can then get x alone, x=\frac {23-y}{y^3-1} We then take the derivative, by the quotient rule. we get the equation of the derivative is
x=-\frac{(y^3-1)+(3y^2)(23-y)}{(y^3-1)^2}.

With our previous equation, x=\frac {23-y}{y^3-1}, we then just sub in y=2, because that's when we want the tangent. we get x=3.

So the tangent passes though (3,2), and, by subbing in y=2 in our derivative, we get the gradient at (3,2) is -37/7 (which is notice is the reciprocal of the value you got). Since y=mx + b we have the gradient of the tangent line is y= (-37/7)x + b. This has to satisfy y=2 and x=3. Therefore we get b = 17 + 6/7.

So my solution for the equation of the tangent line would be : y=-\frac{37}{7}x+ 17\frac{6}{7}

Edit: I just checked my answer on a graphical calculator, looks like i screwed this up badly. Just go through my general method and insturctions, not the specific workings, I think i stuffed up a reciprocal or negative sign...O and btw it looks like my program has gone utterly stupid, it tells me the gradient at that point is zero...eep just Go through my method..