Question

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

Answer 1

De l'Hospital rule applies to undetermined forms like

$\dfrac{0}{0},\quad\dfrac{\infty}{\infty}$

If we evaluate your limit directly, we have

$\displaystyle \lim_{x\to 0}(1+2x)^{-\frac{3}{x}} = 1^\infty$

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

$f(x)=(1+2x)^{-\frac{3}{x}}$

Using the identity $e^{\log(x)}=x$, we can rewrite the function as

$f(x)=e^{\log\left((1+2x)^{-\frac{3}{x}}\right)}$

Using the rule $\log(a^b)=b\log(a)$, we have

$f(x)=e^{-\frac{3}{x}\log(1+2x)}$

Since the exponential function $e^x$ is continuous, we have

$\displaystyle \lim_{x\to 0} e^{f(x)} = e^{\lim_{x\to 0} f(x)}$

In other words, we can focus on the exponent alone to solve the limit. So, we're focusing on

$\displaystyle \lim_{x\to 0} -\frac{3}{x}\log(1+2x)$

Which we can rewrite as

$\displaystyle \lim_{x\to 0} -\frac{3}{x}\log(1+2x) = -3\lim_{x\to 0}\frac{\log(1+2x)}{x}$

Now the limit comes in the form 0/0, so we can apply the theorem: we derive both numerator and denominator to get

$\displaystyle -3\lim_{x\to 0}\frac{\log(1+2x)}{x} = -3 \lim_{x\to 0}\dfrac{\frac{2}{1+2x}}{1} = -3\cdot 2 = -6$

So, the limit of the exponent is -6, which implies that the whole expression tends to

$e^{-6}=\dfrac{1}{e^6}$