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$$\lim_{h\to 0} \frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}=?$$

I evaluated the limit by using the Hopital rule,$$\lim_{h\to 0} \frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}=4^x\lim_{h\to0}\frac{4^h+4^{-h}-2}{h^2}=4^x\lim_{h\to0}\frac{\ln(4)(4^h-4^{-h})}{2h}=4^x\lim_{h\to0}\frac{(\ln4)^2(4^h+4^{-h})}{2}=4^x\times4(\ln2)^2=4^{x+1}(\ln2)^2$$ I want to learn other ideas to solving this problem, so can you please evaluate the limit with other approaches?

Etemon
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    Write the numerator as $4^{x+h}+4^{x-h}-2 \cdot 4^x$ and then see https://math.stackexchange.com/q/1809060/42969 or https://math.stackexchange.com/q/3206176/42969 – Martin R May 15 '22 at 15:19

2 Answers2

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A possible way is using the Taylor expansion $$2\cosh t = e^t+e^{-t} = 2+t^2 + o(t^2)$$

Hence,

\begin{eqnarray*}\frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2} & = & 4^x\frac{e^{h\ln 4}+e^{-h\ln 4}-2}{h^2}\\ & = & 4^x\cdot \frac{2+h^2\ln^2 4 + o(h^2)-2}{h^2} \\ & = & 4^x\cdot\ln^2 4 + o(1) \\ & \stackrel{h\to 0}{\longrightarrow} & 4^x\cdot\ln^2 4 \end{eqnarray*}

trancelocation
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A technique that requires no L'Hopital, Taylor series or anything more complicated that the definition of limit, knowledge of the natural logarithm (e.g. from $\int_1^2\mathrm{d}t/t=\ln 2)$ and continuity of $x\mapsto x^2$:$$\begin{align}4^x\cdot\lim_{h\to0}\frac{4^h+4^{-h}-2}{h^2}&=4^x\cdot\lim_{h\to0}\frac{(4^h-1)^2}{h^24^h}\\&=4^x\cdot\left(\lim_{h\to0}\frac{4^h-1}{h2^h}\right)^2\\&=4^x\cdot4\cdot\left(\lim_{h\to0}\frac{2^h-2^{-h}}{2h}\right)^2\\&=4^{x+1}\cdot(\ln2)^2\end{align}$$

FShrike
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  • Thanks! Can you please elaborate on how you got $\lim_{h\to0}\frac{2^h-2^{-h}}{2h}=(\ln2)^2$ ? – Etemon May 15 '22 at 16:13
  • I didn't. I got: $$\lim_{h\to0}\frac{2^h-2^{-h}}{2h}=\frac{\mathrm{d}}{\mathrm{d}x}\Big|_{x=0}2^x=\ln2$$Instead. Notice the square on the limit in the full post @Etemon

    This is a symmetric derivative. It is equal to the standard derivative when the standard derivative exists.

     – FShrike May 15 '22 at 16:15
  • Oops I wrote the exponent two wrongly... Still I'm not sure which rule you used I know the following rules that connect limits to derivative $f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ and $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$. – Etemon May 15 '22 at 16:21
  • Ok it's clear now. You used $$f'(h)=\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}$$ – Etemon May 15 '22 at 16:26
  • Nice, thanks @Etemon – FShrike May 15 '22 at 16:26