 # Other approaches to evaluate lim_(h => 0) (4^(x+h)+4^(x-h)-4^(x+(1)/(2)))/(h^2) Trevor Rush 2022-08-11 Answered
Other approaches to evaluate $\underset{h\to 0}{lim}\frac{{4}^{x+h}+{4}^{x-h}-{4}^{x+\frac{1}{2}}}{{h}^{2}}$
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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 ${\int }_{1}^{2}\mathrm{d}t/t=\mathrm{ln}2\right)$ and continuity of $x↦{x}^{2}$
$\begin{array}{rl}{4}^{x}\cdot \underset{h\to 0}{lim}\frac{{4}^{h}+{4}^{-h}-2}{{h}^{2}}& ={4}^{x}\cdot \underset{h\to 0}{lim}\frac{\left({4}^{h}-1{\right)}^{2}}{{h}^{2}{4}^{h}}\\ & ={4}^{x}\cdot {\left(\underset{h\to 0}{lim}\frac{{4}^{h}-1}{h{2}^{h}}\right)}^{2}\\ & ={4}^{x}\cdot 4\cdot {\left(\underset{h\to 0}{lim}\frac{{2}^{h}-{2}^{-h}}{2h}\right)}^{2}\\ & ={4}^{x+1}\cdot \left(\mathrm{ln}2{\right)}^{2}\end{array}$
###### Not exactly what you’re looking for? Elisabeth Wiley
A possible way is using the Taylor expansion
$2\mathrm{cosh}t={e}^{t}+{e}^{-t}=2+{t}^{2}+o\left({t}^{2}\right)$
Hence,
$\begin{array}{rcl}\frac{{4}^{x+h}+{4}^{x-h}-{4}^{x+\frac{1}{2}}}{{h}^{2}}& =& {4}^{x}\frac{{e}^{h\mathrm{ln}4}+{e}^{-h\mathrm{ln}4}-2}{{h}^{2}}\\ & =& {4}^{x}\cdot \frac{2+{h}^{2}{\mathrm{ln}}^{2}4+o\left({h}^{2}\right)-2}{{h}^{2}}\\ & =& {4}^{x}\cdot {\mathrm{ln}}^{2}4+o\left(1\right)\\ & \stackrel{h\to 0}{⟶}& {4}^{x}\cdot {\mathrm{ln}}^{2}4\end{array}$