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Find limit as h approaches 0 of $$ \frac{e^{x(h+2)^2} - e^{4x}}{h} $$

I know that somehow this is related to me taking the partial derivative of x

I also know $$\frac{\partial f}{\partial x}(x,y) = \lim_{h\to0} \frac{f(x+h,y) - f(x,y)}{h}$$

But I don't know what to do from here.

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3 Answers 3

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Consider the function $f(x,y) = x y^2 $. Its partial derivative at $y=2$ is given by

$$ \lim_{y \to 2}=\frac{e^{xy^2} - e^{4x}}{y-2}\,.$$

Putting $h=y-2$ transforms the limit to

$$ \lim_{h \to 0}=\frac{e^{x(h+2)^2} - e^{4x}}{h} = \frac{\partial f(x,y)}{\partial y}|_{y=2} =\dots \,. $$

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I would say, this itself wants to be a derivative of some function. What this function could be? Rather write $\frac{\partial f}{\partial y}$ then it will look more similar. Take $y=2$.

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sorry. I'm not sure what you mean. – 40Plot Sep 30 '12 at 23:01

You can expand the square $\frac{e^{x(h+2)^2} - e^{4x}}{h}=\frac{e^{(xh^2+4hx+4x)} - e^{4x}}{h}=\frac{e^{xh^2}e^{4xh}e^{4x}-e^{4x}}h$ then use the Taylor series for the exponentials that have $h$ in them.

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