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Question Number 5705 by sanusihammed last updated on 24/May/16

Answered by FilupSmith last updated on 24/May/16

$$question\left(b\right)$$$$\mathrm{from}\mathrm{previous}\mathrm{answer}\mathrm{in}\mathrm{other}\mathrm{post},$$$$\mathrm{limit}\mathrm{can}\mathrm{be}\mathrm{simplified}\mathrm{to}:$$$$L=\underset{x\to \mathrm{\infty }}{\lim }\frac{3^{2x}-1}{3^{2x}+1}$$$$L=\underset{x\to \mathrm{\infty }}{\lim }\frac{3^{2x}-1}{3^{2x}+1}\times \frac{3^{-2x}}{3^{-2x}}$$$$L=\underset{x\to \mathrm{\infty }}{\lim }\frac{1-3^{-2x}}{1+3^{-2x}}$$$$L=\frac{1-0}{1-0}$$$$\therefore L=1$$

Commented by sanusihammed last updated on 24/May/16

$$Thankssomuch.$$

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