10

Good night guys! I'm having some trouble with this:

$$\lim_{x\to 0} (1+2x)^{1/x}$$

I know that $\lim_{x\to\infty} (1 + 1/x)^x = e$ but I don't know if i should take $h=1/(2x)$ or $h=1/x$

Can someone please help me? Thanks!

Ayman Hourieh
  • 39,603
Mariana Ciotta
  • 331

3 Answers3

16

For problems like these, try taking the natural log of the expression. More precisely, compute $\ln(\lim_{x\rightarrow 0} (1+2x)^{1/x}) = \lim_{x\rightarrow 0} \ln((1+2x)^{1/x}) = \lim_{x\rightarrow 0} \frac{\ln((1+2x))}{x}$. Now it is a form that allows you to apply L'Hopital. Don't for get to exponentiate at the end!

dc2814
  • 1,869
  • wow, but in my book it says that limx→0(1+2x)1/x= e^2 – Mariana Ciotta May 25 '13 at 00:36
  • 3
    Yes, this is true. When you evaluate $\lim_{x\rightarrow 0} \frac{\ln((1+2x))}{x} $ using L'Hopital (taking care about from what direction you are taking the limit) you should get 2. But this is the natural log of the limit you are interested in. Exponentiating, you get that limit you are looking for is $e^2$ – dc2814 May 25 '13 at 00:42
8

We first find the limit as $x$ approaches $0$ from the right.

Let $y=\frac{1}{2x}$. Then $2x=\frac{1}{y}$ and $\frac{1}{x}=2y$. We want $$\lim_{y\to\infty}\left(1+\frac{1}{y}\right)^{2y}.$$ This is the square of the familiar $$\lim_{y\to\infty}\left(1+\frac{1}{y}\right)^{y}.$$ We conclude that $$\lim_{x\to 0^+}\left(1+2x\right)^{1/x}=e^2.$$

Limit from the left is more troublesome with this approach. Let $1+2x=\frac{t}{1+t}$. Then $\frac{1}{x}=-2\frac{1+t}{t}$. So we are looking for $$\lim_{t\to 0^+}\frac{1}{\left(1+t\right)^{-2(1+t)/t}},$$ which is $$ \lim_{t\to 0^+}\left(1+t\right)^{2(1+t)/t}.$$ Now an argument much like the one before works. The limit from the left is also $e^2$.

Vishal Gupta
  • 6,946
André Nicolas
  • 507,029
  • 1
    what was the motivation behind the substitution you took for the evaluation of left hand limit? How were you able to think of such a substitution? – Shoaib Ashraf Dec 16 '19 at 12:13
1

$$\displaystyle\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e,\displaystyle\lim_{x\to-\infty}\left(x+\frac{1}{x}\right)^x=e$$

$(-\infty,0),(0,\infty)\subset\mathbb R-\{0\}=\text{domain}\displaystyle\left(1+\frac{1}{x}\right)^x.$ So $\displaystyle\lim_{x\to0+}(1+x)^\frac{1}{x}=\displaystyle\lim_{x\to0-}(1+x)^\frac{1}{x}=e.$

Now $0$ is a limit point of both $(-\infty,0)$ and $(0,\infty)\implies\displaystyle\lim_{x\to0}(1+x)^\frac{1}{x}=e.$ Let

$$f:\mathbb R-\{0\}\to\mathbb R:x\mapsto2x,\\g:\mathbb R-\{0\}\to\mathbb R:x\mapsto(1+x)^\frac{1}{x}$$

$f(\mathbb R-\{0\})\subset\mathbb R-\{0\}\implies gf:\mathbb R-\{0\}\to\mathbb R:x\mapsto(1+2x)^\frac{1}{2x}$ is defined and $\displaystyle\lim_{x\to0}f(x)=0$ being a limit point of $\mathbb R-\{0\}$ and $\displaystyle\lim_{x\to0}g(x)=e,$ we have $\displaystyle\lim_{x\to0}(1+2x)^\frac{1}{2x}=e.$

Consequently, $\displaystyle\lim_{x\to0}(1+2x)^\frac{1}{2x}=\displaystyle\lim_{x\to0}\left((1+2x)^\frac{1}{2x}\right)^2=e^2.$

Sugata Adhya
  • 3,979