Evaluating $\lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}$

Question

I did this:

$$\begin{align} \lim_{x \to 0} \frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x} &\sim \lim_{x \to 0} \frac{(1 + x + x^2)^{1/x} - (1 + x)^{1/x}}{x} = \\ &= \lim_{x \to 0} \left [ (1+x)^{1/x} \frac{\left ( \frac{1 + x + x^2}{1 + x} \right )^{1/x} - 1}{x} \right ] =\\ &= \lim_{x \to 0} (1+x)^{1/x} \cdot \lim_{x \to 0} \frac{\left ( 1 + \frac{x^2}{1 + x} \right )^{1/x} - 1}{x} =\\ &= e \cdot \lim_{x \to 0} \frac{e^{\frac{1}{x} \cdot \ln \left ( 1 + \frac{x^2}{1+x} \right )} - 1}{x} \sim \\ &\sim e \cdot \lim_{x \to 0} \frac{e^{\frac{x}{1 + x}} - 1}{x} \sim \\ &\sim e \cdot \lim_{x \to 0} \frac{1}{1 + x} = e \end{align}$$

Is it right? If it is, how to evaluate the limit faster? It was pretty long the way I did it.

Answer 1

We have $$\lim_{x\to0}(1+\sin x)^{\dfrac1x}\cdot\lim_{x\to0}\left(1+\frac{\sin^2x}{1+\sin x}\right)^{\dfrac1x}$$

$$\lim_{x\to0}(1+\sin x)^{\dfrac1x}=\left(\lim_{x\to0}(1+\sin x)^{\frac1{\sin x}}\right)^{\lim_{x\to0}\dfrac{\sin x}x}=e^1$$

$$\lim_{x\to0}\left(1+\frac{\sin^2x}{1+\sin x}\right)^{\dfrac1x}=\left(\lim_{x\to0}\left(1+\frac{\sin^2x}{1+\sin x}\right)^{\dfrac{1+\sin x}{\sin^2x}}\right)^{\lim_{x\to0}\dfrac{\sin^2x}{x(1+\sin x)}}=e^0$$

As $\displaystyle\lim_{h\to0}\left(1+h\right)^{\frac1h}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$

Answer 2

I think that, in the same spirit as you show in the post, we could do a little faster considering the development of $$1+\sin (x)+a \sin ^2(x)=1+x+a x^2+O\left(x^3\right)$$ Then going to logarithms, series, followed by exponentiation of series and series again $$f(a)=\Big(1+\sin (x)+a \sin ^2(x)\Big)^{1/x}=e+\left(e a-\frac{e}{2}\right) x+O\left(x^2\right)$$ Then, the numerator is just $$f(1)-f(0) \approx e x-ex^2$$ from which the result follows.

Answer 3

By applying Lagrange's Mean Value Theorem to the function $f(t) = e^t$ we have that $$\begin{align} \frac{1}{x} \left( (1+ \sin x + \sin^2 x)^{1/x} - (1+ \sin x)^{1/x} \right) &= \frac{1}{x} \left( e^{ \frac{ \log(1+ \sin x + \sin^2 x) }{x} } - e^{ \frac{ \log(1+ \sin x) }{x} } \right) \\&= \frac{1}{x} e^{\theta_x} \left( \frac{ \log(1+ \sin x + \sin^2 x)}{x} - \frac{ \log(1+ \sin x)}{x} \right) \\ &= e^{\theta_x} \left( \frac{ \log(1+ \sin x + \sin^2 x) - \log(1+ \sin x)}{x^2} \right) \end{align}$$ where $\theta_x \in \left( \frac{ \log(1+ \sin x) }{x}, \frac{ \log(1+ \sin x + \sin^2 x) }{x} \right)$. Then by squeezing, it is easy to see that $\lim_{x\to0} \theta_x = 1$.

Furthermore we note that by Taylor Series $$\begin{align} \frac{ \log(1+ \sin x + \sin^2 x) - \log(1 + \sin x)}{x^2} &= \frac{ \log(1+ x + x^2 + O(x^3)) - \log(1 + x + O(x^3))}{x^2} \\&= \frac{ x + x^2/2 - ( x - x^2/2) + O(x^3)}{x^2} \\&= 1 + O(x) \end{align}$$

Now passing to the limit as $x\to0$ in the upper expression yields the desired result.

Answer 4

We have $$\begin{aligned}L &= \lim_{x \to 0}\frac{(1 + \sin x + \sin^{2}x)^{1/x} - (1 + \sin x)^{1/x}}{x}\\ &= \lim_{x \to 0}(1 + \sin x)^{1/x}\dfrac{\left(1 + \dfrac{\sin^{2}x}{1 + \sin x}\right)^{1/x} - 1}{x}\\ &= \lim_{x \to 0}e\cdot\dfrac{\left(1 + \dfrac{\sin^{2}x}{1 + \sin x}\right)^{1/x} - 1}{x}\\ &= e\lim_{x \to 0}\dfrac{\left(1 + \dfrac{\sin^{2}x}{1 + \sin x}\right)^{1/x} - 1}{x}\\ &= e\lim_{x \to 0}g(x) = eA\end{aligned}$$ The limit $A = \lim_{x \to 0}g(x)$ can be handled with elementary techniques. Let $$f(x) = \frac{\sin^{2}x}{1 + \sin x}$$ so that $f(x) \to 0$ as $x \to 0$, then we can see that $$\begin{aligned}g(x) &= \frac{(1 + f(x))^{1/x} - 1}{x}\\ &= \dfrac{\exp\left(\dfrac{\log(1 + f(x))}{x}\right) - 1}{x}\\ &= \dfrac{\exp(t) - 1}{x}\end{aligned}$$ Note that $$\begin{aligned}t &= \frac{\log(1 + f(x))}{x}\\ &= \frac{\log(1 + f(x))}{f(x)}\cdot\frac{f(x)}{x}\\ &= \frac{\log(1 + f(x))}{f(x)}\cdot\frac{\sin x}{x}\cdot\frac{\sin x}{1 + \sin x}\\ &\to 1\cdot 1\cdot 0 = 0\end{aligned}$$ Thus we can write $$\begin{aligned}A &= \lim_{x \to 0}g(x)\\ &= \lim_{x \to 0}\frac{e^{t} - 1}{t}\cdot\frac{t}{x}\\ &= \lim_{x \to 0}1\cdot\frac{\log(1 + f(x))}{x^{2}}\\ &= \lim_{x \to 0}\frac{\log (1 + f(x))}{f(x)}\cdot\frac{f(x)}{x^{2}}\\ &= \lim_{x \to 0}1\cdot\frac{\sin^{2}x}{x^{2}(1 + \sin x)}\\ &= \lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{1 + \sin x} = 1\cdot 1 = 1\end{aligned}$$ Thus $L = eA = e\cdot 1 = e$ and the final answer is $e$.

Note: We have used the the limit $(1 + \sin x)^{1/x} \to e$ as $x \to 0$. This is easily handled because $$\log(1 + \sin x)^{1/x} = \frac{\log(1 + \sin x)}{x} = \frac{\log(1 + \sin x)}{\sin x}\cdot\frac{\sin x}{x}\to 1\cdot 1 = 1$$

Answer 5

No the most rigorous, but fast. Define $n=\frac 1x$, and assume it integer (keep in mind the dependency on $x$).

$$\frac{(1 + \sin x + \sin^2 x)^{1/x} - (1 + \sin x)^{1/x}}{x}=\frac{(1 + \sin x + \sin^2 x)^n - (1 + \sin x)^n}{x}.$$

Factoring the numerator as a binomial of the form $a^n-b^n$, you will get two factors:

• the difference of $a-b$, i.e. $\sin^2x$,

• the sum of products of powers $a^{n-1-k}b^k$. All these terms behave like $(1+x+o(x^2))^{n-1}$ and there are $n$ of them. You can recognize the definition of $e$.

So you end up with $$\frac{\sin^2x}xne$$ which tends to $e$.

Answer 6

Recall that \begin{align} \lim_{u\rightarrow0}\Big(1+ua(u)\Big)^{1/u}\xrightarrow{u\rightarrow\infty}e^{a_*}\tag{0}\label{zero} \end{align} if $\lim_{u\rightarrow0}a(u)=a_*$.

On way to check this is by using the inequality $$\frac{y}{1+y}\leq \log(1+y)\leq y, \qquad 1+y>0$$

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Now, \begin{align} \frac1x&\Big((1+\sin x +\sin^{2}x)^{1/x}-(1+\sin x)^{1/x}\Big)\\ &=\frac1x\Big(1+\frac{x\sin x}{x}\Big)^{\frac1x}\Big(\big(1+\tfrac{\sin^2x}{1+\sin x}\big)^{1/x}-1\Big) \end{align}

Let $g(x)=\Big(1+\frac{\sin^2x}{1+\sin x}\Big)^{1/x}$ for $x\neq0$. Notice that $$\Big(1+\frac{\sin^2x}{1+\sin x}\Big)^{1/x}=\Big(1+\frac{x\sin^2x}{x(1+\sin x)}\Big)^{1/x}\xrightarrow{x\rightarrow0}1 $$ by \eqref{zero}. Extend $g$ to $0$ by setting $g(0)=1$. Then

$$\frac{\log\circ g(h)-\log\circ g(0)}{h}=\frac{\log\big(1+\tfrac{\sin^2h}{1+\sin h}\big)}{\tfrac{\sin^2h}{1+\sin h}}\frac{\sin^2h}{h^2(1+\sin h)}\xrightarrow{h\rightarrow0}\log'(1)=1 $$ since $\lim_{k\rightarrow0}\frac{\log(1+k)-\log(1)}{k}=\log'(1)=1$ by definition of derivative of $x\mapsto\log(x)$ at $x=1$. This means that $\log\circ g$ îs differentiable at $x=0$ and so, as $g(x)=\exp\circ(\log\circ g)$, an application of the chain rule yields $$g'(0)=\exp(\log\circ g(0))(\log\circ g)'(0)=e^0\cdot 1=1$$

Similarly, by \eqref{zero} $$\Big(1+\frac{x\sin x}{x}\Big)^{1/x}\xrightarrow{x\rightarrow0}e$$ since $\lim_{x\rightarrow0}\frac{\sin x}{x}=1$. Putting things together we get that $$ \lim_{x\rightarrow0}\frac{(1+\sin x+\sin^2x)^{1/x}-(1+\sin x)^{1/x}}{x}=e $$