Calculate this limit without L'Hopital's rule

Question

I have to calculate this limit whitout using L'Hopital's rule or Taylor polynomials: $$ \lim_{x \to 0} f(x) = (1+\frac{x}{2})^{\frac{1}{x}} $$ I know how to make it using L'Hopital and that the result is $ e^{\frac{1}{2}} $ ,but I'm getting nowhere when I try without it. Any advice?

Answer 1

This is a standard $1^{\infty}$ indeterminate form. A useful result for such limits is: $$ \lim_{x \to a} ~(1+f(x))^{g(x)} = e^{\lim_{x \to a}f(x) \cdot g(x)} ~ \text{if} ~ \lim_{x \to a}f(x) = 0 ~\text{and} \lim_{x \to a}g(x)=\infty $$ In your question, $f(x) = \dfrac{x}{2}$ and $g(x)=\dfrac1x$ $$ \therefore \lim_{x \to a} \left(1+\dfrac{x}{2}\right)^{\frac1x} = e^{\lim_{x \to a} \frac{x}{2}\cdot\frac1x} = e^{\frac12}$$

Answer 2

If $$L=\lim_{x \to a} f(x)^{g(x)}\to 1^{\infty}$$ then $$L=\exp[\lim_{x\to a} [g(x)(f(x)-1))]$$ So here $$L=\exp[\lim_{x\to 0}\frac{1}{x}(1+\frac{x}{2}-1)]=e^{1/2}.$$