This may seem like an easy question, however a few of us are having actually a conflict over it. We space looking in ~ the following limit below, wherein \$ f \$ is a real-valued function on an open subset \$ U \$ that \$ medtox.orgbbR \$ the is differentiable in ~ the point \$ x_0 \$, and where \$ a \$ and also \$ b \$ are in \$ medtox.orgbbR \$:\$\$lim_h o 0 fracf(x + a h) - f(x + b h)h.\$\$

My friend thinks the this is equal to \$ f"(x_0) \$, but I don’t see just how to prove that this is true.

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Thanks because that the help!

The easiest method to prove your friend dorn is to use this come a function that we really understand, choose \$f(x) = x\$. We"ll do it in ~ \$x_0 = 1\$. Then you are calculating

\$\$ fracf(1 + ah) - f(1 + bh)h = frac1 + ah - 1 - bhh = (a - b).\$\$

This is not \$1\$ and depends critically top top \$a\$ and \$b\$ (exactly choose derivatives do not).

More generally, we have the right to write (but whereby I omit all border signs)\$\$ eginalignfracf(x_0 + ah) - f(x_0 + bh)h &= fracf(x_0 + ah) - f(x_0)h - fracf(x_0 + bh) - f(x_0)h \& o af"(x_0) - bf"(x_0) = (a-b)f"(x_0).endalign\$\$

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