## Tuesday, April 28, 2015

### Calculate $\displaystyle \lim_{x \to \infty} (\sin \sqrt{x+1}-\sin \sqrt{x})$

Calculate $\displaystyle \lim_{x \to \infty} (\sin \sqrt{x+1}-\sin \sqrt{x})$.

Let:

[MATH]L=\lim_{x\to\infty}\left(\sin\left(\sqrt{x+1} \right)-\sin\left(\sqrt{x} \right) \right)[/MATH]

Application of the sum-to-product identity:

[MATH]\sin(\alpha)-\sin(\beta)=2\sin\left(\frac{\alpha-\beta}{2} \right)\cos\left(\frac{\alpha+\beta}{2} \right)[/MATH]

And the limit property:

[MATH]\lim_{x\to c}\left(k\cdot f(x) \right)=k\cdot\lim_{x\to c}\left(f(x) \right)[/MATH]

Allows us to write:

[MATH]L=2\lim_{x\to\infty}\left(\sin\left(\frac{\sqrt{x+1}-\sqrt{x}}{2} \right)\cos\left(\frac{\sqrt{x+1}+\sqrt{x}}{2} \right) \right)[/MATH]

Rationalization of the numerator of the sine function gives us:

[MATH]L=2\lim_{x\to\infty} \left( \sin \left(\frac{1}{2 \left( \sqrt{x+1}+ \sqrt{x} \right)} \right) \cos \left(\frac{ \sqrt{x+1}+ \sqrt{x}}{2} \right) \right)[/MATH]

Now, using the property of limits:

[MATH]\lim_{x\to c}\left(f(x)\cdot g(x) \right)=\left(\lim_{x\to c}\left(f(x) \right) \right)\left(\lim_{x\to c}\left(g(x) \right) \right)[/MATH]

We obtain:

[MATH]L=2 \lim_{x \to\infty} \left( \sin \left( \frac{1}{2 \left( \sqrt{x+1}+ \sqrt{x} \right)} \right) \right) \lim_{x \to\infty} \left( \cos \left( \frac{ \sqrt{x+1}+ \sqrt{x}}{2} \right) \right)[/MATH]

The first limit goes to zero, and the second limit is bounded, hence:

[MATH]L=0[/MATH]