Find the limit of: $\lim_{x\to0}(\frac{x^2+2}{x^2})^{\sin(x)}$

Question

I strugle finding the $\lim_{x\to0}(\frac{x^2+2}{x^2})^{\sin(x)}$. I wanted to say that: let $1+t = \frac{x^2+2}{x^2}$, which gives me $x^2=\frac{2}{t}$ and there I don't know how to go any further. Until now, all the problems of that sort were of degree 1, and you could simply put $t$ in, play a bit with the exponentials and then get a solution. If it would be $\cos(x)$, I could say that $\cos(-x) = \cos(x)$, but that's unfortunately not true for $\sin(x)$.

I then tried to say that $\sin(x)=\sqrt{1-\cos^2(x)}$, which gives me after a few operations: $\lim_{t\to0}(1+t)^{\sqrt{\cos^2(\frac{\sqrt{2}}{\sqrt{t}}})})$, but I don't know how to continue, neither do I know whether it's correct.

Answer 1

Hint

Start with $y=\left(\frac{x^2+2}{x^2}\right)^{\sin x}$, then \begin{align*} \ln y & = \sin x \ln\left(\frac{x^2+2}{x^2}\right)\\ & = \sin x \left[\ln (x^2+2) -2\ln(x)\right]\\ & = \sin x \left[\ln (x^2+2)\right] -2\sin x \left[\ln(x)\right]. \end{align*} as $x \to 0$, the first term goes to $0$. So you now have to focus on the second term and write it as $$2\frac{\sin x}{x}x \ln (x).$$ Now use the fact that $\frac{\sin x}{x} \to 1$ as $x \to 0$. The other expression $x \ln x$ is easy to deal with using L'Hospital.

Answer 2

Hint: For small $x$, $\sin x \approx x$.

Answer 3

Let $L =\lim_{x\to 0}\left(\frac{x^2+2}{x^2}\right)^{\sin x}$. Then

\begin{align} \ln L &= \lim_{x\to 0} \left(\sin x \ln\left(1+\frac2{x^2}\right)\right)\\ &= \lim_{x\to 0}\left(\frac{\sin x}x\cdot x\ln\left(1+\frac2{x^2}\right)\right)\\ &= \lim_{x\to 0} \left(x\ln\left(1+\frac2{x^2}\right)\right)\\ &= \lim_{x\to 0} \frac{\ln\left(1+\frac2{x^2}\right)}{\frac 1x}\\ \end{align}

L'Hospital from here gives $\ln L = 0\implies L = 1$.

Answer 4

With asymptotic analysis, it's not so hard. First note it suffices to find the limit of the log. Then $$\ln\Bigl(\frac{x^2+2}{x^2}\Bigr)^{\sin x}=\sin x\ln(x^2+2)-2\sin x \ln x$$ The first term tends to $0$. As to the second term, we know $\;\sin x\sim_0x$, hence $$2\sin x\ln x\sim_0 2x\ln x\xrightarrow[x\to 0]{}0,$$ so that $\;\ln\Bigl(\dfrac{x^2+2}{x^2}\Bigr)^{\sin x}\!\!\!\xrightarrow[x\to 0]{}0 $ and finally $$\lim_{x\to 0}\Bigl(\dfrac{x^2+2}{x^2}\Bigr)^{\sin x}=1.$$