# How to approach problems on limits

1. Dec 15, 2006

### 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

2. Dec 15, 2006

### helpm3pl3ase

anyone????????

3. Dec 16, 2006

### Gib Z

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

We'll im 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 thats 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 thats 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..

Last edited: Dec 16, 2006
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