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Does anyone know how to evaluate the following limit?

$$ f'(-2)=\lim_{h\to0}\frac{(-2+h)e^{-2+h}+2e^{-2}}{{h}} $$

What is $f(x)$ ?

I want to see a step by step solution if possible.

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    $\begingroup$ $f(x)=xe^x$ by the definition of a derivative... Is that what you mean by "What is the value of $f(x)$?" $\endgroup$ – Shahar Aug 25 '14 at 15:17
  • $\begingroup$ -e^x + any constant. INfact there must be other condition be given to tell f(x) but still if only this much information is given you are left with nothing just place f'(x) replacing -2 with x $\endgroup$ – DSinghvi Aug 25 '14 at 16:52
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Let's evaluate our limit.

$$ \begin{align} & \lim_{h \to 0^+} \dfrac{(-2 + h) e^{-2+h} + 2e^{-2}}{h} = \\ = & \lim_{h \to 0^+} \dfrac{-2e^{-2}(e^h - 1)}{h} + \lim_{h \to 0^+} \dfrac{he^{-2}}{h} = \\ = & -2e^{-2} \underbrace{\lim_{h \to 0^+} \dfrac{(e^h - 1)}{h}}_{= 1} + e^{-2} \underbrace{\lim_{h \to 0^+} \dfrac{h}{h}}_{= 1} = \\ = & -e^{-2} \end{align} $$

About the question How is f(x), we can suppose, since $f'(-2) = -e^{-2}$, that $f'(x) = -e^{x}$. Then, we deduce that $f(x) = -e^{x} + \alpha$, with $\alpha \in \mathbb{R}$.

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    $\begingroup$ did't you plug in a wrong variable -x which sould be +x @unomoinverde $\endgroup$ – DSinghvi Aug 25 '14 at 17:51
  • $\begingroup$ Yeah, you're right. I'll fix it. Thanks! $\endgroup$ – Filippo De Bortoli Aug 25 '14 at 18:05
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HINT

take 2e^-2 common and separate he^-2 now since both limits exist when divided by h separate them , You'll find it

Final answer will come f(x) = -e^x + a constant

Though there can be other function possible also . for that read my comment

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