SOLUTION: Having trouble understanding how limits questions like this work, could you please explain how to solve this type of question?
find lim h->0 (f(x+h)-f(x))/h for f(x)=sqrt(7+2x)
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Question 600349: Having trouble understanding how limits questions like this work, could you please explain how to solve this type of question?
find lim h->0 (f(x+h)-f(x))/h for f(x)=sqrt(7+2x)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
As ugly as this thing looks, the solution is both simple and elegant.
}\ -\ \sqrt{7\ +\ 2x}}{h})
The trick is to multiply by 1, but in a special form. Remember the factorization of the difference of two squares? Go back the other way: The product of a conjugate pair of binomials, i.e. (a\ +\ b))
is the difference of two squares, 
. So what? Watch:
Multiply by 1 in the form of }\ +\ \sqrt{7\ +\ 2x}}{\sqrt{7\ +\ 2(x\ +\ h)}\ +\ \sqrt{7\ +\ 2x}})
}\ -\ \sqrt{7\ +\ 2x}}{h}\right)\left(\frac{\sqrt{7\ +\ 2(x\ +\ h)}\ +\ \sqrt{7\ +\ 2x}}{\sqrt{7\ +\ 2(x\ +\ h)}\ +\ \sqrt{7\ +\ 2x}}\right))
Now create the difference of two squares in the numerator:
\ -\ (7\ +\ 2x)}{h\left(\sqrt{7\ +\ 2(x\ +\ h)}\ +\ \sqrt{7\ +\ 2x}\right)})
Collect like terms in the numerator:
}\ +\ \sqrt{7\ +\ 2x}\right)})
Get rid of the factor of 
that is now in both numerator and denominator.
}\ +\ \sqrt{7\ +\ 2x}})
Now take the limit
making the other just go away.
Collect terms in the denominator
And finally, lose the 2 that is in both numerator and denominator:
Done.
John
My calculator said it, I believe it, that settles it
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