4
$\begingroup$

fine the limits-without-lhopital rule

$$\lim_{ x \to0^- }\frac{2^{\frac{1}{x}}+2^{\frac{-1}{x}}}{3^{\frac{1}{x}}+3^{\frac{-1}{x}}}=?$$

My Try :

$h= \frac{1}{x} :h\to - \infty$

so :

$$\lim_{ h\to - \infty }\frac{2^{h}+2^{-h}}{3^{h}+3^{-h}}=?\\\lim_{ h\to - \infty}\frac{(2^{-h})2^{2h}+1}{(3^{-h})3^{2h}+1}=?\\\lim_{ h\to - \infty }\frac{(2^{-h})2^{2h}+1}{(3^{-h})3^{2h}+1}=?$$

now :?

  • 7
    $$\lim_{h\to -\infty}\left(\frac{2}{3}\right)^{-h}\frac{2^{2h}+1}{3^{2h}+1}=0\cdot 1=0$$2017-02-28

1 Answers 1

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$$\frac{2^{\frac{1}{x}}+2^{\frac{-1}{x}}}{3^{\frac{1}{x}}+3^{\frac{-1}{x}}}=\left(\frac32\right)^{1/x}\frac{2^{2/x}+1}{3^{2/x}+1}\xrightarrow[x\to0^-]{}0\cdot\frac{0+1}{0+1}=0$$