Since [tex]\cos(\pi+0)+1=\cos(\pi)+1=-1+1=0[/tex], the limit take on the indeterminate form 0/0. So, we can use the L'Hôpital's Rule:
[tex]\lim_{h\to0}\dfrac{\cos(\pi+h)+1}{h}=\lim_{h\to0}\dfrac{\sin(\pi+h)+0}{1}\\\\
\lim_{h\to0}\dfrac{\cos(\pi+h)+1}{h}=\lim_{h\to0}\sin(\pi+h)\\\\
\lim_{h\to0}\dfrac{\cos(\pi+h)+1}{h}=\sin(\pi+0)\\\\
\lim_{h\to0}\dfrac{\cos(\pi+h)+1}{h}=\sin(\pi)\\\\
\boxed{\lim_{h\to0}\dfrac{\cos(\pi+h)+1}{h}=0}[/tex]
