Limit of $\left(2-a^\frac{1}{x}\right)^x$ How do I prove the following limit?
$$\lim_{x \to \infty}\left(2-a^\frac{1}{x}\right)^x = \frac{1}{a}$$
 A: Let 
$$L= \lim_{x\to\infty}(2-a^{\frac{1}{x}})^x$$
so passing $\ln$ in the above equation we get
$$\ln L= \ln(\lim_{x\to\infty}(2-a^{\frac{1}{x}})^x)=
\lim_{x\to\infty}\ln((2-a^{\frac{1}{x}})^x)=
\lim_{x\to\infty}x\ln (2-a^{\frac{1}{x}})=
\lim_{x\to\infty}\frac{\ln (2-a^{\frac{1}{x}})}{\frac{1}{x}}$$
Using L'hopital's rule we get
$$\ln L=\lim_{x\to\infty}\frac{\frac{-a^{\frac{1}{x}}(\frac{-1}{x^2})\ln a}{2-a^{\frac{1}{x}}}}{\frac{-1}{x^2}}= \lim_{x\to\infty} \frac{-a^{\frac{1}{x}}\ln a}{2-a^{\frac{1}{x}}}=-\ln a= \ln a^{-1}$$
Therefore
$$L=a^{-1} \therefore  \lim_{x\to\infty}(2-a^{\frac{1}{x}})^x=a^{-1}=\frac{1}{a}$$
A: $$\lim_{x \to \infty}\left(2-a^\frac{1}{x}\right)^x = \lim_{x \to \infty} e^{x \ln(2-a^{\frac{1}{x}})}=\lim_{x \to \infty} e^{x \ln(1+1-a^{\frac{1}{x}})}=\lim_{x \to \infty} e^{x \ln(1-(a^{\frac{1}{x}}-1))}=\lim_{x \to \infty} e^{x \ln(1-\frac{1}{x}\ln a)}=$$
$$= \lim_{x \to \infty} e^{x \ln(1-\frac{1}{x}\ln a)}= \lim_{x \to \infty} e^{x (-\frac{1}{x}\ln a)}=\lim_{x \to \infty} e^{-\ln a}=e^{\ln(\frac{1}{a})}=\frac{1}{a}$$
A: $$\lim_{x \to \infty} \left(2-a^\frac{1}{x}\right)^x 
= \exp \log \lim_{x \to \infty} \left(2-a^\frac{1}{x}\right)^x
= \exp \lim_{x \to \infty} \log \left(2-a^\frac{1}{x}\right)^x
= \exp \lim_{x \to \infty} x \log \left(2-a^\frac{1}{x}\right)
= \exp \lim_{t \searrow 0} \frac{1}{t} \log \left(2-a^t\right)
= \exp \lim_{t \searrow 0} \frac{\log 2-a^t}{t}
$$
Using L'Hospital:
$$
= \exp \lim_{t \searrow 0} \frac{\overbrace{a^t}^{\to 1} \log(a)}{\underbrace{a^t}_{\to 1}-2}
= \exp \frac{1 \log(a)}{1-2}
= \exp (- \log(a))
= \frac{1}{a}
$$
