Limits?

Hey everyone, Ok I have this one question where I can not find the slope of the tangent using limits.

This is the function:

1 / root(x)

the point is:

( 4, 1/2)

I need to find the slope of the tangent using that and then find the equation which I can do if I had the slope! I have tried for couple of hours right now and no luck anyone can help! I seem to be either stuck somehwere or did something wrong!

Thanks again for your help!

y = 1/sqrt{x} = x^(-1/2)

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

x = 4, dy/dx = -(1/2)(1/8) = -1/16 which is gradient of slope

Finding the gradient through fist principles (limits) will require the use of the binomial expansion to expand (x + h)^(-1/2)

Ok that is great the only problem is that we are using the two calculus formulas:

which are:

1. m = lim(x -> a) f(x) - f(a)/x - a

and

2. m = lim(h -> 0) f(a+h) - f(a)/h

When I try with them it is very confusing, anyone can help me out becase I need an answer to bring tomorrow.

Quote:

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Originally Posted by Glaysher

Finding the gradient through fist principles (limits) will require the use of the binomial expansion to expand (x + h)^(-1/2)

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Wait what do you mean expand by multiply it or to the power of (-1/2)???

Anyone can help me out please???

f(x) = x^-1/2

Using L to mean limit (h->0),

f'(x) = L [f(x+h) - f(x)] / h = L [(x+h)^-1/2 - x^-1/2] / h] = L [(x^1/2 - (x+h)^1/2) / (hx^1/2(x+h)^1/2]

Now we multiply both the numerator and the denominator by [x^1/2 + (x+h)^1/2]

f'(x) = L {[x^1/2 - (x+h)^1/2]*[x^1/2 + (x+h)^1/2] / [hx^1/2(x+h)^1/2]*[x^1/2+(x+h)^1/2]} = L {[x - x - h] / [hx^1/2(x+h)^1/2]*[x^1/2 + (x+h)^1/2]} = (now replace h by 0) = -1/(x*2x^1/2) = (-1/2)x^-3/2

At x=4, the slope is +1/16 and -1/16. But the slope is to be calculated at (4, 1/2) so it's the positive branch. Therefore, slope = -1/16