Calculate $\lim_{x \longrightarrow 0^{-}}(1-2^x)^{\sin(x)}$ by Taylor expansion.

Question

Calculate the limit by Taylor expansion:(in a step I don't know what to do?)

$\lim_{x \longrightarrow 0^{-}}(1-2^x)^{\sin(x)}$

solution

$y=(1-2^x)^{\sin x} \longrightarrow \ln y=(\sin x)\ln(1-2^x)$

$\lim_{x \longrightarrow 0^{-}}\ln y = \lim_{x \longrightarrow 0^{-}}(\sin x)\ln(1-2^x)=$

In this step I don't know what done

$=\lim_{x \longrightarrow 0^{-}}(\sin x)\ln(1-e^{x\ln 2})$

$=\lim_{x \longrightarrow 0^{-}}x\ln(1-(1+x\ln 2))$

$=\lim_{x \longrightarrow 0^{-}}x\ln(-x\ln 2)$

$=\lim_{x \longrightarrow 0^{-}}x\ln(-x)+\lim_{x \longrightarrow 0^{-}}x\ln(\ln 2)$

$=\lim_{x \longrightarrow 0^{-}}\frac{\ln(-x)}{\frac{1}{x}}$

$=\lim_{x \longrightarrow 0^{-}}\frac{\frac{-1}{-x}}{\frac{-1}{x^2}}$

$=\lim_{x \longrightarrow 0^{-}}(-x) = 0 \Longrightarrow$

$\Longrightarrow \lim_{x \longrightarrow 0^{-}}\ln y = 0 \Longrightarrow$

$\lim_{x \longrightarrow 0^{-}}y = e^0 = 1 \Longrightarrow$

$\lim_{x \longrightarrow 0^{-}}(1-2^x)^{\sin x} = 1$

Answer 1

Note that

$$(1-2^x)^{\sin(x)}=e^{\sin(x)\log(1-2^x)}\to 1$$

indeed

$${\sin(x)\log(1-2^x)}={\frac{\sin(x)}{1-2^x}(1-2^x)\log(1-2^x)}\to 0$$

since by standard limits

• $(1-2^x)\log(1-2^x)\to 0$
• $\frac{\sin(x)}{1-2^x}=\frac{\sin(x)}{x}\frac{x}{1-2^x}\to1\cdot\frac1{-\log 2}=-\frac1{\log 2}$