Two ways come to mind.
The first involves simply invoking the definition of the derivative at a point. If it exists, then it's equal to
[tex]f'(c)=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}[/tex]
where [tex]h=x-c\iff x=c+h[/tex]. This means the limit in question is given by
[tex]\displaystyle\lim_{h\to0}\frac{\tan\left(\frac\pi4+h\right)-1}h=\lim_{x\to\pi/4}\frac{\tan x-\tan\frac\pi4}{x-\frac\pi4}=\lim_{x\to\pi/4}\frac{\tan x-1}{x-\frac\pi4}[/tex]
i.e. the derivative of [tex]f(x)=\tan x[/tex] at [tex]x=\dfrac\pi4[/tex]. We have
[tex]f'(x)=\sec^2x[/tex]
so the value of the limit is
[tex]f'\left(\dfrac\pi4\right)=\sec^2\dfrac\pi4=(\sqrt2)^2=2[/tex]
The other way to go about this is with trig identities. I seem to always forget the sum identity for tangent, but that's easily derivable if you know the corresponding (co)sine identities.
[tex]\tan\left(\dfrac\pi4+h\right)=\dfrac{\sin\left(\frac\pi4+h\right)}{\cos\left(\frac\pi4+h\right)}=\dfrac{\sin\frac\pi4\cos h+\cos\frac\pi4\sin h}{\cos\frac\pi4\cos h-\sin\frac\pi4\sin h}=\dfrac{\tan\frac\pi4+\tan h}{1-\tan\frac\pi4\tan h}[/tex]
[tex]\implies\tan\left(\dfrac\pi4+h\right)=\dfrac{1+\tan h}{1-\tan h}[/tex]
Now,
[tex]\displaystyle\lim_{h\to0}\frac{\tan\left(\frac\pi4+h}-1}h=\lim_{h\to0}\frac{\frac{1+\tan h}{1-\tan h}-1}h[/tex]
[tex]=\displaystyle\lim_{h\to0}\frac{(1+\tan h)-(1-\tan h)}{h(1-\tan h)}[/tex]
[tex]=\displaystyle2\lim_{h\to0}\frac{\tan h}{h(1-\tan h)}[/tex]
[tex]=2\displaystyle\left(\lim_{h\to0}\frac{\sin h}h\right)\left(\lim_{h\to0}\frac1{\cos h(1-\tan h)}\right)[/tex]
[tex]=2[/tex]
