If we have the following function $$\phi(x)=\frac{e^{m_2x}-e^{m_1x}}{m_2-m_1}$$ and we want to find the following limit $$\lim_{m2\to m_1}{\frac{e^{m_2x}-e^{m_1x}}{m_2-m_1}}$$ where $$m_2=m_1+h$$ then I think it must be $$\frac{d}{dm}e^{mx}|_{m=m_1}=e^{m_1x}x$$ My first question:
Can we substitute by m2 instead of m1 ? or we must substitute by m1 because m2 tends to m1 ? I mean can we write that the above limit will result in : $$\frac{d}{dm}e^{mx}|_{m=m_2}=e^{m_2x}x$$? I tried to prove that we can substitute by m2 also , I proved it but I am worried that we can not do that or that I may have proved it in a wrong way: $$\lim_{m2\to m_1}{e^{m_2x}\frac{1-e^{(m_1-m_2)x}}{-(m_1-m_2)}}$$ $$\lim_{h\to 0}{e^{m_2x}\frac{1-e^{hx}}{-h}}$$ using l'hopital rule $$\lim_{h\to 0}{e^{m_2x}\frac{e^{-hx}x}{-1}}=xe^{m_2x}$$
(this is my first question)
My second question: what about the following limit ( if we let m1 tends to m2 instead) $$\lim_{m1\to m_2}{\frac{e^{m_2x}-e^{m_1x}}{m_2-m_1}},\ \ m_2=m_1+h$$ will it result in $$\frac{d}{dm}e^{mx}|_{m=m_2}=e^{m_2x}x$$?
I like the following way: $$\lim_{m_2\rightarrow m_1}\frac{e^{m_2x}-e^{m_1x}}{m_2-m_1}=x\lim_{m_2\rightarrow m_1}\frac{e^{m_1x}\left(e^{(m_2-m_1)x}-1\right)}{(m_2-m_1)x}=xe^{m_1x}.$$