limh→0 f((6+h)−f(6)) / h ,when f(x)= √x -4

f(6)=√6-4. (6+h)-f(6)=6+h-(√6-4)=6-√6+4+h=10-√6+h.

f((6+h)-f(6))/h=f(10-√6+h)/h=(√(10-√6+h)-4)/h=???

I think the question should have been written:

lim h➝0 (f(6+h)-f(6)) / h, which makes better sense.

So we have (√(6+h)-4-(√6-4))/h=(√(6+h)-√6-4+4)/h=(√(6+h)-√6)/h.

We can write √(6+h) as √(6(1+(h/6)))=√6(1+(h/6))^½.

This approximates to √6(1+h/12)=√6+h√6/12 when h is small.

So (√(6+h)-√6)/h=(√6+h√6/12-√6)/h=√6/12 which can also be written √6/(2×6)=1/(2√6). This evaluates to 0.2041 approximately.

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