This is calculus using L'hospitals rule Lim x --> 0 (1+2x)^(-3/x)

Accepted Solution

De l'Hospital rule applies to undetermined forms like[tex]\dfrac{0}{0},\quad\dfrac{\infty}{\infty}[/tex]If we evaluate your limit directly, we have[tex]\displaystyle \lim_{x\to 0}(1+2x)^{-\frac{3}{x}} = 1^\infty[/tex]which is neither of the two forms covered by the theorem.So, in order to apply it, we need to write the limit as follows: we start with[tex]f(x)=(1+2x)^{-\frac{3}{x}}[/tex]Using the identity [tex]e^{\log(x)}=x[/tex], we can rewrite the function as[tex]f(x)=e^{\log\left((1+2x)^{-\frac{3}{x}}\right)}[/tex]Using the rule [tex]\log(a^b)=b\log(a)[/tex], we have[tex]f(x)=e^{-\frac{3}{x}\log(1+2x)}[/tex]Since the exponential function [tex]e^x[/tex] is continuous, we have[tex]\displaystyle \lim_{x\to 0} e^{f(x)} = e^{\lim_{x\to 0} f(x)}[/tex]In other words, we can focus on the exponent alone to solve the limit. So, we're focusing on[tex]\displaystyle \lim_{x\to 0} -\frac{3}{x}\log(1+2x) [/tex]Which we can rewrite as[tex]\displaystyle \lim_{x\to 0} -\frac{3}{x}\log(1+2x) = -3\lim_{x\to 0}\frac{\log(1+2x)}{x}[/tex]Now the limit comes in the form 0/0, so we can apply the theorem: we derive both numerator and denominator to get[tex]\displaystyle -3\lim_{x\to 0}\frac{\log(1+2x)}{x} = -3 \lim_{x\to 0}\dfrac{\frac{2}{1+2x}}{1} = -3\cdot 2 = -6[/tex]So, the limit of the exponent is -6, which implies that the whole expression tends to[tex]e^{-6}=\dfrac{1}{e^6}[/tex]