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Here we have an indeterminacy:

$\lim_{h \to 0}\dfrac{e^{-x}- e^{-x-h}}{h} = \dfrac{e^{-x}- e^{-x-0}}{0} = \dfrac{0}{0}.$

So we can use L'Hopital's rule differentiating each function: $\dfrac{d}{dh} \left(e^{-x} - e^{-x-h}\right) = \dfrac{d}{dh} (e^{-x}) - \dfrac{d}{dh} (e^{-x-h}) = 0 + e^{-x-h} = e^{-x-h}.$

Is it possible to get the same result without using differentiation? If possible, could you demonstrate?

amWhy
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rafaelcb21
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1 Answers1

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We can use the following : Given $f$ a differentiable function, by definition $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f^\prime(x)$$ Now take $f(x)=e^{-x}$. Then the above becomes $$\lim_{h\to 0}\frac{e^{-x-h}-e^{-x}}{h}=(e^{-x})^\prime=-e^{-x}$$ Then multiplying everything with $-1$, we get $$\lim_{h\to 0}\frac{e^{-x}-e^{-x-h}}{h}=e^{-x}$$ which is the limit desired.

Chanel Rose
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