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I am given the piecewise function $G(t)$, which is written below, and I am asked to find all values of the parameters $\alpha$ and $\beta$ for which the function $G$ is differentiable at $t=1$.

$G(t)=\alpha t^{2}+\ln(t)$ if $t\geq 1$

$G(t)=\beta e^{t-1}-2t$ if $t<1$

I am also given the hint:

You may assume that:$$\lim_{x\to0} \frac{e^{x}-1}{x}=1$$ $$\lim_{x\to 0} \frac{\ln(1+x)}{x}=1$$

I know that the function must be continuous, so $\lim\limits_{t\to 1^+}G(t)=\lim\limits_{t\to 1^-}G(t)$. This leaves me with $\beta-2=\alpha$. I am a bit confused with my next steps. I believe I must use the definition of a derivative in order to set the slopes of each parts of $G(t)$ to be equal to each other. We are not far enough in the course to simply take the derivatives of each part of $G(t)$.

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For the function to be differentiable at $1$, you need $$\lim_{x\to 0}\frac{G(1+x)-G(1)}{x}$$ to exist. Note that $G(1) = \alpha$.

Evaluating the limit from the right, we have $$\begin{align*} \lim_{x\to 0^+}\frac{G(1+x)-G(1)}{x} &= \lim_{x\to 0^+}\frac{\alpha(1+x)^2+\ln(1+x)-\alpha }{x}\\ &= \lim_{x\to 0^+}\frac{\alpha\bigl((1-x)^2-1\bigr) + \ln(1+x)}{x}\\ &= \lim_{x\to 0^+}\frac{\alpha(x^2-2x)}{x} + \lim_{x\to 0^+}\frac{\ln(1+x)}{x}\\ &=\lim_{x\to 0+}\alpha(x-2) + 1\\ &= 1-2\alpha. \end{align*}$$

Now try evaluating the limit from the left in a similar manner; the two limits must be equal for the derivative to exist. (If the derivative exists, this automatically guarantees continuity).

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It worked out very well, thank you! –  Malthus Oct 20 '11 at 14:56
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Your belief is true, the definition of the derivative is in order here:

$ G'(t) = \lim_{h\rightarrow 0}\frac{G(t+h)-G(t)}{h} $

And in your case, $t=1$, so you need to evaluate $G(1)$ and $G(1+h)$ for the right-end limit and $G(1-h)$ for the left-end limit of $G'(t)$ (just as you did for the continuity part). When you evaluate

$ \lim_{h\rightarrow 0}\frac{G(t+h)-G(t)}{h} $

you'll see that the two fractions given in the "you may assume" part will pop up and you can use those assumptions to help you with the limit.

I can clarify more if needed.

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