# Limit of a trigonometric rational expression

How to evaluate the limit of this expression?

$$\lim_{x\to\infty} \frac{\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}$$

I managed to simplify the denominator into a sinus form by the Pythagorean formula, and also modified the argument of the sinus in the numerator by dividing with a conjugate which I think is necessary, into this final form:

$$\lim_{x\to\infty} \frac{\sin^2\left(\frac{1}{\sqrt{x+1}+\sqrt{x}}\right)}{\sin^2\frac{1}{x}}$$

And that's where I got stuck. So far I've tried to come up with some solution to eliminate the sinus from the denominator, without success. I'm sure the solution is pretty obvious but I don't quite know how to modify the trigonometric functions in this fashion, so any help would be most appreciated.

Now, setting $t=\dfrac{1}{\sqrt{x+1}+\sqrt{x}}$ we note $x\to\infty$ implies $t\to 0^+$, and since $\displaystyle{\lim_{t\to 0}\frac{\sin t}{t}=1}$ $$\lim_{x\to\infty}\frac{\sin^2(\sqrt{x+1}-\sqrt{x})}{(\sqrt{x+1}-\sqrt{x})^2}=\left[\lim_{x\to \infty}\frac{\sin \frac{1}{\sqrt{x+1}+\sqrt{x}}}{\frac{1}{\sqrt{x+1}+\sqrt{x}}}\right]^2=\left(\lim_{t\to 0^+}\frac{\sin t}{t}\right)^2=(1)^2=1$$ Similarly, putting $t=1/x$ we have $t\to0^+$ as $x\to\infty$, and $$\lim_{x\to\infty}\frac{\sin^2(1/x)}{1/x^2}=\left(\lim_{t\to 0^+}\frac{\sin t}{t}\right)^2=1^2=1$$ Then, \begin{align} \lim_{x\to\infty}\frac{\sin^2\left(\sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}&=\left(\lim_{x\to \infty}\frac{\left(\sqrt{x+1}-\sqrt{x}\right)^2}{\frac{1}{x^2}}\right)\lim_{x\to\infty}\frac{\frac{\sin^2\left(\sqrt{x+1}-\sqrt{x}\right)}{(\sqrt{x+1}-\sqrt{x})^2}}{\frac{\sin^2(1/x)}{1/x^2}}\\[5pt] &=\left(\lim_{x\to\infty}x^2\left(\frac{1}{\sqrt{x+1}+\sqrt{x}}\right)^2\right)\cdot\frac{1}{1}\\ &=\lim_{x\to\infty}\left(\frac{x}{\sqrt{x+1}+\sqrt{x}}\right)^2 \end{align} You can determine the last limit.
First lets use $\sin^2{x}=\frac{1-\cos{2x}}{2}$ to turn both expressions to linear expressions in cosine. Then you can use L'Hospital's rule.
You could also solve this limit using the Taylor expansion of $\sin(x)=x-x^3/3!+x^5/5!\mp \cdots$ combined with the squeez theorem.