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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 $$\frac{d}{dm}e^{mx}|_{m=m_2}=e^{m_2x}x$$?

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  • $\begingroup$ I think, you are right. $\endgroup$ – Michael Rozenberg Nov 21 '17 at 3:16
  • $\begingroup$ I edited the post $\endgroup$ – MCS Nov 21 '17 at 3:54
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I like the following way: $$\lim_{m_2\rightarrow m_1}\frac{e^{m_2x}-e^{m_1x}}{m_2-m_1}=x\lim_{m_2\rightarrow m_1}\frac{e^{m_1x}\left(e^{(m_2-m_1)x}-1\right)}{(m_2-m_1)x}=xe^{m_1x}.$$

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  • $\begingroup$ I do not understand what did you do after taking the exponential common factor .. did you apply l'hopital rule considering (m2-m1)x as a single variable? if yes , I think we do not need to multiply and divide by x , we can take only m2-m1 as a single variable .. right ? ( I am just checking whether i understand ) $\endgroup$ – MCS Nov 21 '17 at 3:58
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    $\begingroup$ @Sousa I used $\lim\limits_{x\rightarrow0}\frac{e^x-1}{x}=1.$ $\endgroup$ – Michael Rozenberg Nov 21 '17 at 4:01
  • $\begingroup$ Note : I edited the post , I proved that we can substitute by m2 instead of m1 ..However, I feel that we can NOT substitute by m2 since m2 tends to m1 , so the proof may be wrong and I would like to know where is the fault in the proof i wrote . Thanks . $\endgroup$ – MCS Nov 21 '17 at 4:03
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    $\begingroup$ I think, $m_2\rightarrow m_1$ says that $m_2$ is changed and $m_1$ is a constant. $\endgroup$ – Michael Rozenberg Nov 21 '17 at 4:07

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