Verified answer

lt h→0

(√2+h - √2) / h

Multiply and divide by (√2+h + √2)

= (√2+h - √2)*(√2+h + √2) / h*(√2+h + √2)

= (2+h - 2) / h*(√2+h + √2)

= (h) / h*(√2+h + √2)

= 1/(√2+h + √2)

Now put lt h→0

= 1/(√2+0 + √2)

= 1/2√2

enable f(h) = [sqrt(2) + h - sqrt(2)]/h = h/h At h = 0, that's undefined. yet, lim h ->0 [h/h] = a million there's a great distinction between putting the cost of 0 in a function and comparing the decrease with the aid of fact the function has a tendency to 0. in the 1st case, i could no longer cancel out the h's with the aid of fact if h became equivalent to 0, it would've been an invalid step. yet in a decrease, you are able to assume that h isn't 0 yet very on the verge of fall down of 0. for this reason i ought to cancel it out to get the decrease as a million.

Lim h→ 0 ( (√2+h - √2 ) / h)

= Lim h→ 0 (2 + h - 2 ) / h(√2+h + √2 )

= Lim h→ 0 h / [h(√2+h + √2 )]

= Lim h→ 0 1/(√2+h + √2 )

= 1/(√2 + √2)

= 1/2√2

= √(1/8)

Well in terms of calculus, you get 0/0 so you use a certain rule where you have to take the derivative of the equation so the answer would be 1

(sqrt(h + 2) - sqrt(2)) / h * (sqrt(h + 2) + sqrt(2)) / (sqrt(h + 2) / sqrt(2))

= (h + 2 - 2) / h(sqrt(h + 2) + sqrt(2))

= h / h(sqrt(h + 2) + sqrt(2))

= 1 / (sqrt(h + 2) + sqrt(2))

Now let h = 0:

= 1 / (sqrt(2) + sqrt(2))

= 1 / 2sqrt(2)

= 1 / sqrt(8)

= sqrt(8) / 8

= sqrt(2) / 4

= 0.354 (to 3 decimal places).

So yes, you are correct.

(√2 + h - √2 ) / h

. . . √2 - √2 = 0

h / h ... remember here that h is "approaching" 0 ... not "equal" to 0

. . . h/h = 1

1 <=== NOW let h = 0

1 <=== answer