# $\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$

Calculate: $$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$$

I don't know how to use L'Hôpital's Rule.

I tried to make $\tan x =\frac{\sin x}{\cos x}$ for the term ${\sqrt{\tan x}}$.

• Maybe this will help: L'Hospital's Rule. – Pichi Wuana Mar 7 '16 at 23:05
• It's Forbidden to use it – user315918 Mar 7 '16 at 23:07
• Are you permitted to use Taylor Series ? – WW1 Mar 7 '16 at 23:15

You can first remove a few factors $$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}\\ =\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt{\tan x}}{x \sqrt{x}}\lim_{x \to 0^+}\frac{1-\cos x\sqrt{\cos x}}{x^2}\\ =\lim_{x \to 0^+}\frac{1-\cos x\sqrt{\cos x}}{x^2}.$$ Then multiply by the conjugate $$=\lim_{x \to 0^+}\frac{1-\cos^3 x}{x^2(1+\cos x\sqrt{\cos x})},$$ evaluate the finite factor at denominator $$=\frac12\lim_{x \to 0^+}\frac{1-\cos x(1-\sin^2 x)}{x^2},$$ use trigonometric identitites $$=\frac12\lim_{x \to 0^+}\frac{2\sin^2\frac x2+\cos x\sin^2 x}{x^2},$$ and conclude $$=\left(\frac12\right)^2+\frac12.$$

We used

$$\frac{\tan x}x=\frac{\sin x}x\frac1{\cos x}\to 1,$$ $$\frac{\sin ax}x=a\frac{\sin ax}{ax}\to a.$$

• $\lim_{x\to0}\frac{1-\cos^3x}{x^2}=\lim_{x\to0}\frac{1-\cos x}{x^2}(1+\cos x+\cos^2x)=\lim_{x\to0}\frac{1-\cos^2x}{x^2}\frac{1+\cos x+\cos^2x}{1+\cos x}=\frac{3}{2}$ – egreg Mar 8 '16 at 0:12

\begin{align} \lim_{x\to0^{+}}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}} &=\lim_{x\to0^{+}}\frac{\tan^{3/2}x-\sin^{3/2} x}{x^3 \sqrt{x}}\\ &=\lim_{x\to0^{+}}\frac{\sin^{3/2}x}{x^{3/2}}\frac{\sec^{3/2}x-1}{x^2}\\ &=\lim_{x\to0^{+}}\left(\frac{\sin x}{x}\right)^{3/2}\cdot \lim_{n\to\infty}\frac{\sec^{3/2}x-1}{x^2}\\ &=\lim_{x\to0^{+}}\frac{\sec^{3/2}x-1}{x^2}\\ &=\lim_{x\to0^{+}}\frac{\left(1+\frac12x^2+\frac1{24}x^4+\cdots\right)^{3/2}-1}{x^2}\\ &=\lim_{x\to0^{+}}\frac{\left(1+\frac34x^2+\cdots\right)-1}{x^2}\\ &=\frac34 \end{align}

• What's $sec$ means ? – user315918 Mar 7 '16 at 23:24
• @user315918 $\,\sec x\;$ = secant of $\;x=\frac1{\cos x}\;$ – DonAntonio Mar 7 '16 at 23:32

$$\frac{\tan x\sqrt{\tan x}-\sin x\sqrt{\sin x}}{x^3\sqrt x}=\left(\frac{\sin x} x\right)^{3/2}\cdot\frac{\frac1{\cos^{3/2}x}-1}{x^2}=$$

$$=\left(\frac{\sin x} x\right)^{3/2}\frac{1-\cos^{3/2}x}{x^2\cos^{3/2}x}=\left(\frac{\sin x} x\right)^{3/2}\frac{1-\cos^2x+\cos^{3/2}x(\cos^{1/2}x-1)}{x^2\cos^{3/2}x}=$$

$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2+\frac{\cos^{1/2}x-1}{x^2}\right]=$$

$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2+\frac{\cos x-1}{x^2(\cos^{1/2}x+1)}\right]=$$

$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2-\frac2{\cos^{1/2}x+1}\frac{\sin^2\frac x2}{x^2}\right]=$$

$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2-\frac2{\cos^{1/2}x+1}\cdot\frac14\cdot\left(\frac{\sin\frac x2}{\frac x2}\right)^2\right]\xrightarrow[x\to0^+]{}$$

$$\rightarrow1\left[1\cdot1-\frac22\cdot\frac14\cdot1\right]=1-\frac14=\frac34$$