Find a limit without using L'Hopitals rule 9

Question

Can someone please show me how to do this without using L'Hopitals rule:

$$\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x$$

I know the limit is $e^a$, but I would like to know the steps taken to get to that answer.

thank you!

Answer 1

$y=(1+a/x)^x$, $\ln y = x \ln(1+a/x)= x(a/x-(a/x)^2/2+\cdots)\approx a$ when $x\to \infty$, then $y =e^a$

Answer 2

Assuming that you defined $x \mapsto e^x$ as the inverse of $\ln$, which was defined by $$ \ln(x) = \int_1^x \frac{1}{t} dt, $$ you know that $\ln$ is continuous on the positive reals, and maps them onto the whole real line.

Because $$ \lim_{x \to b} f(g(x)) =f(\lim_{x \to b} g(x) $$ when $f$ is continuous everywhere, you can take the log of the limit that you're looking for. \begin{align} L &= \lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x\\ \ln L &= \ln(\lim_{x \to \infty} \left(1 + \frac{a}{x}\right)^x)\\ \ln L &= \lim_{x \to \infty} \ln(\left(1 + \frac{a}{x}\right)^x)\\ \ln L &= \lim_{x \to \infty} x\ln\left(1 + \frac{a}{x}\right)\\ \end{align} Letting $y = x/a$, this becomes \begin{align} \ln L &= \lim_{x \to \infty} x\ln\left(1 + \frac{a}{x}\right)\\ \ln L &= \lim_{y \to \infty} ay\ln\left(1 + \frac{1}{y}\right)\\ \ln L &= \lim_{y \to \infty} ay\ln\left(\frac{y+1}{y}\right)\\ \ln L &= a\lim_{y \to \infty} y\left(\ln(y+1) - \ln(y)\right)\\ \end{align} The expression in parentheses is just $$\int_y^{y+1} \frac{1}{t} dt$$, which is between $\frac{1}{y+1}$ and $\frac{1}{y}$, so the whole expression in the limit lies between $\frac{y}{y+1}$ and $\frac{y}{y}$. By the squeeze lemma, the limit must be one, so the right-hand side is $a$. So $$ \ln L = a\\ L = \exp(a). $$

Answer 3

Here we will use a method of substitution which is motivated by the definition of $e$ $$ e = \lim_{n \to \infty} \left(1 + \frac{1}{n} \right)^n$$ We want to "get rid" of the $a$ which we find in the equation. To do first fix $a \not = 0$. Now we can see that if $a > 0$ a substitution of the form $$ m = \frac{n}{a}$$ will do the job! Indeed we have $$ m \to \infty \ \text{as} \ n \to \infty$$ and furthermore $n = ma$ $$ \lim_{n \to \infty} \left(1 + \frac{a}{n} \right)^n = \lim_{m \to \infty} \left(1 + \frac{a}{ma} \right)^{(ma)} = \lim_{m \to \infty} \left(\left(1 + \frac{1}{m} \right)^m\right)^a$$ Now we are able to use the formal definition of $e$ and we note that when the limit exists the power of a limiting equation is just the power of the limit so we have $$ \lim_{m \to \infty} \left(\left(1 + \frac{1}{m} \right)^m\right)^{a} = e^a$$ Now the more interesting case is $a < 0$. Now we need to consider the transformation $$ m = - \frac{n}{a}$$ again we have $$ m \to \infty \ \text{as} \ n \to \infty$$ However let us observe what happens with the substitution this time. I will skip some of the similar parts and get to the intriguing part! We are left with $$ \lim_{m \to \infty} \left(1 - \frac{1}{m} \right)^{-ma} $$ This seemingly intractable equation transforms very nicely consider $$ \left(1 - \frac{1}{m} \right)^{-m} = \left(\frac{m - 1}{m}\right)^{-m} = \left(\frac{m}{m - 1}\right)^m = \left(\frac{m + 1 - 1}{m - 1}\right)^m = \left( 1 + \frac{1}{m - 1}\right)^{m}$$ The next substitution almost fell into my lap! So finally consider $$ p = m - 1$$ Clearly $$ p \to \infty \ \text{as} \ m \to \infty$$ So we will get $$ \lim_{m \to \infty} \left(1 - \frac{1}{m} \right)^{-ma} = \lim_{p \to \infty} \left(\left(1 + \frac{1}{p} \right)^{p+1}\right)^{a} = \lim_{p \to \infty} \left(\left(1 + \frac{1}{p} \right)^{p}\right)^{a} \left(1 + \frac{1}{p}\right)^{a} = e^a \cdot 1$$