I’m going to assume that f(x)=e²ˣ, so f(0.05)=1.10517...
df/dx=2e²ˣ, at x=a=0, df/dx=2, so the gradient for the linear approximation is 2. f(a)=f(0)=1.
When x=0.05, L(0.05)=1+0.1=1.1.
Error is (f(0.05)-1.1)/f(0.05)=0.00468 or 0.47%.
[If I assume f(x)=e^x², df/dx=2xe^x²=0 when a=0, so L(x) would have no gradient, it would be a horizontal line. That means that the above assumption is probably correct.]