<s>[INST] let $f(x) = \left\lbrace\begin{array}{ll} x^2+3x, &x \geq1\\ x^2-3x+6, & x<1 \end{array}\right.$ then find the limit… Let 
$f(x) =
\begin{cases}
x^2+3x,  &x \geq1\\[2ex]
x^2-3x+6, & x<1
\end{cases}$
Then find the
$$\lim_{h \to0} \frac{f(1)-f(1-h^2)}{h^2}=?$$

My Try :
$$f(1)=4$$
$$f(1-h^2)=(1-h^2)^2-3(1-h^2)+6=h^4-h^2+4=h^2(h^2-1)+4$$
So we have :
$$\lim_{h \to0} \frac{h^2(h^2-1)}{h^2}=-1$$
it is right ?
 [/INST] It should be observed that the limit in question is left hand derivative of $f$ at $1$ and hence it is equal to $2x-3$ evaluated at $x=1$. Thus the desired limit is $-1$. It should be also observed that the above works because $f$ is continuous at $1$.
</s>