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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Tristian Clayton

Answered 2022-08-12
Author has **12** answers

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 ${\int}_{1}^{2}\mathrm{d}t/t=\mathrm{ln}2)$ and continuity of $x\mapsto {x}^{2}$

$\begin{array}{rl}{4}^{x}\cdot \underset{h\to 0}{lim}\frac{{4}^{h}+{4}^{-h}-2}{{h}^{2}}& ={4}^{x}\cdot \underset{h\to 0}{lim}\frac{({4}^{h}-1{)}^{2}}{{h}^{2}{4}^{h}}\\ & ={4}^{x}\cdot {\left(\underset{h\to 0}{lim}\frac{{4}^{h}-1}{h{2}^{h}}\right)}^{2}\\ & ={4}^{x}\cdot 4\cdot {\left(\underset{h\to 0}{lim}\frac{{2}^{h}-{2}^{-h}}{2h}\right)}^{2}\\ & ={4}^{x+1}\cdot (\mathrm{ln}2{)}^{2}\end{array}$

$\begin{array}{rl}{4}^{x}\cdot \underset{h\to 0}{lim}\frac{{4}^{h}+{4}^{-h}-2}{{h}^{2}}& ={4}^{x}\cdot \underset{h\to 0}{lim}\frac{({4}^{h}-1{)}^{2}}{{h}^{2}{4}^{h}}\\ & ={4}^{x}\cdot {\left(\underset{h\to 0}{lim}\frac{{4}^{h}-1}{h{2}^{h}}\right)}^{2}\\ & ={4}^{x}\cdot 4\cdot {\left(\underset{h\to 0}{lim}\frac{{2}^{h}-{2}^{-h}}{2h}\right)}^{2}\\ & ={4}^{x+1}\cdot (\mathrm{ln}2{)}^{2}\end{array}$

Elisabeth Wiley

Answered 2022-08-13
Author has **1** answers

A possible way is using the Taylor expansion

$2\mathrm{cosh}t={e}^{t}+{e}^{-t}=2+{t}^{2}+o({t}^{2})$

Hence,

$\begin{array}{rcl}\frac{{4}^{x+h}+{4}^{x-h}-{4}^{x+\frac{1}{2}}}{{h}^{2}}& =& {4}^{x}\frac{{e}^{h\mathrm{ln}4}+{e}^{-h\mathrm{ln}4}-2}{{h}^{2}}\\ & =& {4}^{x}\cdot \frac{2+{h}^{2}{\mathrm{ln}}^{2}4+o({h}^{2})-2}{{h}^{2}}\\ & =& {4}^{x}\cdot {\mathrm{ln}}^{2}4+o(1)\\ & \stackrel{h\to 0}{\u27f6}& {4}^{x}\cdot {\mathrm{ln}}^{2}4\end{array}$

$2\mathrm{cosh}t={e}^{t}+{e}^{-t}=2+{t}^{2}+o({t}^{2})$

Hence,

$\begin{array}{rcl}\frac{{4}^{x+h}+{4}^{x-h}-{4}^{x+\frac{1}{2}}}{{h}^{2}}& =& {4}^{x}\frac{{e}^{h\mathrm{ln}4}+{e}^{-h\mathrm{ln}4}-2}{{h}^{2}}\\ & =& {4}^{x}\cdot \frac{2+{h}^{2}{\mathrm{ln}}^{2}4+o({h}^{2})-2}{{h}^{2}}\\ & =& {4}^{x}\cdot {\mathrm{ln}}^{2}4+o(1)\\ & \stackrel{h\to 0}{\u27f6}& {4}^{x}\cdot {\mathrm{ln}}^{2}4\end{array}$

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