Limit of exponential and logarithm function

Question

Question:

Find $\lim_{x\to\infty} \left(1-\frac{e}{x}\right)^{x^2}$

Attempt:

We can solve this limit by using the natural logarithm and the limit definition of the exponential function. Let's begin by taking the natural logarithm of the expression inside the limit:

$\ln\left[\left(1-\frac{e}{x}\right)^{x^2}\right] = x^2 \ln\left(1-\frac{e}{x}\right)$

Now we can rewrite the limit as:

$\lim_{x\to\infty} x^2 \ln\left(1-\frac{e}{x}\right)$

Let's focus on the expression inside the natural logarithm. We can use the Taylor series expansion for $\ln(1-x)$:

$\ln\left(1-\frac{e}{x}\right) = -\frac{e}{x} - \frac{e^2}{2x^2} - \frac{e^3}{3x^3} - \cdots$

As $x$ approaches infinity, all the terms after the first one become negligible, so we can approximate the expression as:

$\ln\left(1-\frac{e}{x}\right) \approx -\frac{e}{x}$

Substituting this back into our limit, we get:

$\lim_{x\to\infty} x^2 \ln\left(1-\frac{e}{x}\right) \approx \lim_{x\to\infty} -e x = -\infty$

Finally, we can take the exponential functio n of both sides to get the original limit:

$\lim_{x\to\infty} \left(1-\frac{e}{x}\right)^{x^2} = e^{-\infty} = \boxed{0}$

Therefore, the limit is 0.

Is this correct ?

Answer 1

As already commented, your proof is absolutely blameless but alternatively, you could have used $\lim_{u\to0}\left(1+u\right)^{1/u}=e,$ which instantly gives: $$\lim_{x\to^+\infty}\left(1-\frac ex\right)^{x^2}=\lim_{x\to+\infty}\left(\left(1-\frac ex\right)^{-\frac xe}\right)^{-ex}=e^{-\infty}=0.$$

Your approximation of $\ln\left(1-\frac ex\right)$ by $-\frac ex$ as $x\to\infty$ is classical but it does not need Taylor series, and should be denoted by $\sim$ instead of $\approx:$ $$\ln(1+u)\sim u\quad\text{as}\quad u\to0,$$ which has a precise meaning: $$\lim_{u\to0}\frac{\ln(1+u)}u=1.$$ (No need for L'Hospital for this limit, it is simply the derivative at $1$ of $\ln,$ by definition.)

And the asymptotic equivalence is especially designed to be perfectly compatible

• with products, i.e. if (as $x\to\infty$ for instance) $f_1(x)\sim g_1(x)$ and $f_2(x)\sim g_2(x),$ then $f_1(x)f_2(x)\sim g_1(x)g_2(x);$
• with limits, i.e. if (as $x\to\infty$ for instance) $f(x)\sim g(x)$ and $g(x)\to\ell$ then $f(x)\to\ell.$

Answer 2

Let $n=\lfloor{x}\rfloor.$ Then $$\left (1-{e\over x}\right)^{x^2}\le \left ( 1-{1\over n+1}\right )^{n^2}={1\over \left ( {1+{1\over n}}\right )^{n^2}}$$ By applying the binomial formula we obtain $$\left ( {1+{1\over n}}\right )^{n^2}\ge 1+{n^2\over n}\ge x$$ Summarizing we get $$\left (1-{e\over x}\right)^{x^2}\le {1\over x}$$

Answer 3

Your proof is basically correct but the asymptotic approximation can be made tighter as follows: $$x^2((-\frac{e}{x})-\ln(1-\frac{e}{x}))=\frac{e^2}{2}+\frac{e^3}{3x}+\frac{e^4}{4x^2}+...$$ $$<\frac{e^2}{2}+\frac{e^3}{2x}+\frac{e^4}{2x^2}+...$$ This last expression is just a geometric series with initial term $\frac{e^2}{2}$ and ratio $\frac{e}{x}$ If $x>0$ and the ratio <1 then the geometric sertes converges to $$\frac{\frac{e^2}{2}}{1-\frac{e}{x}}$$ which, if $x \ge 4$ is less than $$\frac{\frac{e^2}{2}}{1-\frac{3}{4}}=2e^2.$$ Thus$$\vert x^2\ln(1-\frac{e}{x})-x^2(-\frac{-e}{x}) \vert<2e^2$$ so $$-2e^2-ex<x^2\ln(1-\frac{e}{x})<2e^2-ex$$ Alternavely, you could take the limit of $$\frac{\ln(1-\frac{e}{x})}{\frac{1}{x^2}}$$ as$x \to \infty$ by l'Hospital's Rule.

Answer 4

One more way to solve: $$ \lim_{x \to \infty}{(1-\frac{e}{x})^{x^2}} = \\ \lim_{x \to \infty}{\exp(\ln((1-\frac{e}{x})^{x^2}))} = \\ \exp(\lim_{x \to \infty}{\ln((1-\frac{e}{x})^{x^2})}) = \\ \exp(\lim_{x \to \infty}{x^2\ln(1-\frac{e}{x})}) = \\ \exp(\lim_{x \to \infty}{x^2(-\frac{e}{x})\frac{(\ln(1-\frac{e}{x}) - 0)}{-\frac{e}{x}}}) = \\ \exp(\lim_{x \to \infty}{x^2(-\frac{e}{x})\frac{(\ln(1-\frac{e}{x}) - \ln(1)}{-\frac{e}{x}}}) = \\ \exp(\lim_{x \to \infty}{(x^2(-\frac{e}{x}))}\lim_{x \to \infty}{\frac{(\ln(1-\frac{e}{x}) - \ln(1)}{-\frac{e}{x}}}) = \\ \exp(\lim_{x \to \infty}{(x^2(-\frac{e}{x}))}\ln'(1)) = \\ \lim_{x \to \infty}{e^{-ex}} = 0$$