How to show $\lim\limits_{t\to 0}\frac{|f+tg|^{p}-|f|^{p}}{t}=pfg|f|^{p-2}$ Could anyone help to compute this limit? Thank you!
This is part of a proof in Analysis by E. Lieb. Let $f$ and $g$ be real numbers and $p>1$, show the following limit:
$$\lim\limits_{t\to 0}\frac{|f+tg|^{p}-|f|^{p}}{t}=pfg|f|^{p-2}$$
 A: If $g=0$, both sides are constant $0$.
If $g\neq 0$ and $f=0$, we have
$$\lim_{t\to 0}\frac{|tg|^p}{t} = |g|^p\lim_{t\to 0}\frac{|t|^p}{t} = 0,$$
since $p\gt 1$ (it's the limit of $|g|t^{p-1}$ as $t\to 0^+$, which is $0$, and of $-|g|t^{p-1}$ as $t\to 0^-$, which is also $0$). 
If $g\neq 0$ and $f\neq 0$, say $f\gt 0$. Using L'Hopital's Rule you have
$$\begin{align*}
\lim_{t\to 0}\frac{|f+tg|^p - |f|^p}{t} &= \lim_{t\to 0}\frac{(f+tg)^p - f^p}{t}\\
&= \lim_{t\to 0}\frac{p(f+tg)^{p-1}g - 0}{1} = pf^{p-1}g = pfgf^{p-2} = pfg|f|^{p-2}.
\end{align*}$$
If $f\lt 0$, then
$$\begin{align*}
\lim_{t\to 0}\frac{|f+tg|^p - |f|^p}{t} &= \lim_{t\to 0}\frac{(-f-tg)^p - (-f)^p}{t}\\
&= \lim_{t\to 0}\frac{p(-f-tg)^{p-1}(-g) - 0}{1} \\
&= p(-f)^{p-1}(-g)\\
&= -pg|f|^{p-1}\\
&= p(-|f|)g|f|^{p-2}\\
&= pfg|f|^{p-2}.
\end{align*}$$
In any of the four cases, we have equality.
(As to getting rid of the absolute value signs in $|f+tg|$, so long as $f\neq 0$, $f+tg$ has the same sign as $f$ for sufficiently small $t$).
A: The given limit is easily seen to be the (definition of the) derivative of the function $$ t \mapsto |f+gt|^p $$ at $t = 0$. We can calculate this derivative using the chain rule. This comes out to be
$$
p |f+gt|^{p-1} \times \frac{f+gt}{|f+gt|} \times g.
$$
Here, the first factor is the derivative of the function $z \mapsto z^p$ at $|f+gt|$. The second factor is the derivative of the absolute value function at $f+gt$.  Upon substituting $t=0$, we get the claimed result. 
To make the above proof watertight, observe that the absolute value function is differentiable only when the argument is nonzero; so in the above, we have to assume that $f \neq 0$. One would then need to complete the proof by discussing the case $f=0$ separately, as in @Arturo Magidin's answer. 
