As H goes to Zero, what's the limit of .... *CALCULUS*?

Question

5/X-2 +3

so i have y prime = 5 / (x+h-2) + 3 ( x+h-2) / 1 (x+h-2) =

5 + 3 (x+h-2) / (x+h-2) .... im so lost after this !!! please help i'm in calclus I and i have no idea what's going on .

Answer 1

f(x) = 5/(x-2) + 3



df/dx = lim(h->0) [f(x+h) - f(x)]/h



f(x+h) = 5/(x+h-2) + 3



f(x+h) - f(x) = 5/(x+h-2) + 3 - 5/(x-2) - 3 = 5/(x+h-2) - 5/(x-2)



= [5(x-2) - 5(x+h-2)]/(x+h-2)(x-2)



= [5x - 10 - 5x - 5h + 10]/(x+h-2)(x-2)



= -5h/(x+h-2)(x-2)



Using that result for f(x+h) - f(x) in the limit, the h in the numerator cancels the h in the denominator and we have:



df/dx = lim(h->0) [f(x+h) - f(x)]/h = lim(h->0) -5/(x+h-2)(x-2) = -5/(x-2)²

Answer 2

you do no longer explicitly state what g(x) is, and with the equation appearing extremely confusing, it quite is tough to make certain what it quite is. pointing out that g(a million) = 2 would not define the function. The reasoning at the back of allowing a value to bypass to a particular decrease to that end h tending to 0, is to make certain the by-product and as a result the tangent of a function initially from an approximation in the direction of an exact outcome. the only way i think of i ought to help is to grant an occasion with an undemanding function to artwork with. think f(x) = x² Now a tangent to this function at a particular element c say, might nicely be approximated be drawing a right now line crossing the graph of f(x) at 2 factors f(a) and f(b), such that a < c < b The slope of the line by way of those factors is given by potential of the elementary outcome (f(b) - f(a)) / (b - a) ie (b² - a²) / (b - a) Now bear in mind, it quite is an approximation to the tangent of the function at f(c). the greater suitable the era h = b - a, the greater severe the approximation; the smaller the era, the greater suitable the approximation. It follows then that this approximation gets greater suitable using fact the era gets smaller till finally, we've an exact tangent to the function. it would help to visualize this by potential of attempting some sketches of the graph of x² and drawing lines between nearer and nearer factors approximately some intermediate element c². using fact the above argument is actual for any cost of x, we are in a position to place a = x and positioned b = x + h and decrease the era between x and x + h. to that end the slope of the line between f(x) and f(x + h) is (f(x + h) - f(x)) / ((x + h) - x) ie - with f(x) = x² the slope m of this line is m = ((x + h)² - x²) / ((x + h) - x) = ((x² + 2xh + h²) - x²) / h = ( 2xh + h²) / h = (2x + h) Letting h?0, we get the by-product df/dx = 2x So, the tangent of the function f(x) = x² at x to that end the tangent of the function at f(c) = c² is m = 2c to make certain the equation of the line at this element, we would desire to discover the y-intercept ok say. it quite is, we would desire to discover ok such that y(0) = ok all of us be attentive to that the slope of the line at c² is 2c so as that y = 2cx + ok however the slope variety of the equation is (c² - ok) / (c - 0) = 2c c² - ok = 2c² ok = - c² So the equation of the line is y = 2cx - c² on the element (c, c²) although the above occasion became into for representation, it would grant help to in determining the by-made of g(x) on the element (a million, g(a million)) - ie (a million, 2)

Answer 3

remember that a fraction only equals zero if the numerator is equal to zero. so set it up so that

5+3(x+h-2)=0 then solve for x. once you find x (it will contain an h), substitute that into the first equation for the x variable. Then, simply plug in the number 0 for h.

Answer 4

What you have to do is to treat h as being equal to 0.

Also, 3 is just a number and when you differentiate that, it just goes to 0.

So your initial set up is going to be:



(5/((x+h)-2)-5/(x-2))/h



This should help give you a head start.

Answer 5

are you trying to evaluate the difference quotient for f(x)=5/(x-2)+3?

(f(x+h)-f(x))/h as h approaches 0



(5/(x+h-2)+3 - (5/(x-2)+3))/h

(5/(x+h-2)-5/(x-2))/h

(5(x-2)/((x-2)(x+h-2))-5(x+h-2)/((x+h-2)(x-2)))/h

((5x-10-5x-5h+10)/(x^2-2x+hx-2h-2x+4))/h



-5h/(h(x^2-4x+hx-2h+4))

-5/(x^2-4x+hx-2h+4)

lim h-->0

-5/(x^2-4x+4)

-5/(x-2)^2