4
$\begingroup$

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 ${\sqrt{\tan x}}$.

$\endgroup$
3
  • $\begingroup$ Maybe this will help: L'Hospital's Rule. $\endgroup$ – Pichi Wuana Mar 7 '16 at 23:05
  • $\begingroup$ It's Forbidden to use it $\endgroup$ – user315918 Mar 7 '16 at 23:07
  • $\begingroup$ Are you permitted to use Taylor Series ? $\endgroup$ – WW1 Mar 7 '16 at 23:15
4
$\begingroup$

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

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

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

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

$\endgroup$
2
  • $\begingroup$ What's $sec$ means ? $\endgroup$ – user315918 Mar 7 '16 at 23:24
  • 1
    $\begingroup$ @user315918 $\,\sec x\;$ = secant of $\;x=\frac1{\cos x}\;$ $\endgroup$ – DonAntonio Mar 7 '16 at 23:32
0
$\begingroup$

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

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

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

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

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

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.