This exercise $$\lim_{x\to 0+}\left[\frac{1}{x^\frac{3}{2}}-\frac{1}{x^\frac{1}{2}\sin(x)}\right]$$ was in my calculus III test at college, after trying hard to solve it I was not really able to do so, I know intuitively that this limit equals 0 but did not find an appropriate way to prove it, as the limit gives an indeterminate form $\infty – \infty$. I worked out the fractions and got $$\lim_{x\to 0+}\frac{x^\frac{1}{2}\sin(x) – x^\frac{3}{2} }{x^2\sin(x)}$$ which gives me an indeterminate form $\frac{0}{0}$, then using L'hopitals rule, the limit starts getting uglier as more indeterminate forms keep showing up. My college test is over, but, still I want to find out how to solve this limit, any help would be highly appreciated.
Prove that $\lim_{x\to 0+}\left[\frac{1}{x^{3/2}}-\frac{1}{x^{1/2}\sin(x)}\right]=0$
calculuslimitstrigonometry
Best Answer
I wouldn't rationalise the denominator. Instead, stick with $$\frac{\sin x - x}{x^{\frac{3}{2}}\sin x}.$$ Applying L'Hopital's rule, \begin{align*} \frac{\cos x - 1}{\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}}\cos x} &= \frac{(\cos x - 1)(\cos x + 1)}{\left(\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}}\cos x\right)(\cos x + 1)} \\ &= \frac{-\sin^2 x}{\left(\frac{3}{2}x^{\frac{1}{2}}\sin x + x^{\frac{3}{2}}\cos x\right)(\cos x + 1)} \\ &= \frac{-x^{\frac{1}{2}}\frac{\sin^2 x}{x^2}}{\left(\frac{3}{2}\frac{\sin x}{x} + \cos x\right)(\cos x + 1)} \\ &\to \frac{-0 \cdot 1}{\left(\frac{3}{2} \cdot 1 + 1\right)(1 + 1)} = 0. \end{align*}