Differentiation (Part-38) | S N De

10. Trigonometric and Inverse Trigonometric Limits

$10(i)$ Evaluate: $$\lim_{x \to 0}\frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin^2x}$$

Solution

Let $\cos x = z$. As $x \to 0$, $z \to 1$.

$$\lim_{x \to 0}\frac{\sqrt{\cos x}-\sqrt[3]{\cos x}}{\sin^2x} = \lim_{z \to 1}\frac{z^{\frac{1}{2}}-z^{\frac{1}{3}}}{1-z^2} \quad [\because \sin^2x = 1-\cos^2x = 1-z^2]$$ $$= \lim_{z \to 1}\left[-\frac{(z^{\frac{1}{2}}-z^{\frac{1}{3}})(z+z^{\frac{1}{2}}\cdot z^{\frac{1}{3}}+z^{\frac{2}{3}})}{(z+1)(z-1)(z+z^{\frac{1}{2}}\cdot z^{\frac{1}{3}}+z^{\frac{2}{3}})}\right]$$ $$= -1 \times \lim_{z \to 1} \frac{(z^{1/2})^3-(z^{1/3})^3}{(z+1)(\sqrt{z}+1)(\sqrt{z}-1)(z+z^{\frac{1}{2}}\cdot z^{\frac{1}{3}}+z^{\frac{2}{3}})}$$ $$= -1 \times \lim_{z \to 1}\frac{z^{3/2}-z}{(z+1)(\sqrt{z}+1)(\sqrt{z}-1)(z+z^{5/6}+z^{2/3})}$$ $$= -1 \times \lim_{z \to 1} \frac{z(\sqrt{z}-1)}{(z+1)(\sqrt{z}+1)(\sqrt{z}-1)(z+z^{5/6}+z^{2/3})}$$ $$= -1 \times \lim_{z \to 1}\frac{z}{(z+1)(\sqrt{z}+1)(z+z^{5/6}+z^{2/3})}$$ $$= \frac{-1 \times 1}{(1+1)(\sqrt{1}+1)(1+1^{5/6}+1^{2/3})} = \frac{-1}{2 \times 2 \times 3} = -\frac{1}{12} \quad \text{(Ans.)}$$

$10(ii)$ Evaluate: $$\lim_{x \to 0}\frac{1-\cos x}{x^2}$$

Solution

$$\lim_{x \to 0}\frac{1-\cos x}{x^2} = \lim_{x \to 0} \frac{2\sin^2(x/2)}{x^2}$$ $$= \frac{1}{2} \times \lim_{x \to 0} \frac{4 \sin^2(x/2)}{x^2} = \frac{1}{2} \times \lim_{x \to 0} \frac{\sin^2(x/2)}{(x/2)^2}$$

Substituting $z = x/2$, as $x \to 0$, $z \to 0$:

$$= \frac{1}{2} \times \lim_{z \to 0} \frac{\sin^2 z}{z^2} = \frac{1}{2} \times \left(\lim_{z \to 0} \frac{\sin z}{z}\right)^2 = \frac{1}{2} \times 1^2 = \frac{1}{2} \quad \text{(Ans.)}$$

$10(iii)$ Evaluate: $$\lim_{h \to 0}\frac{1-\cos h}{h \sin h}$$

Solution

Dividing both numerator and denominator by $h^2$:

$$\lim_{h \to 0}\frac{1-\cos h}{h \sin h} = \lim_{h \to 0}\frac{(1-\cos h)/h^2}{\frac{h\sin h}{h^2}}$$ $$= \frac{\lim_{h \to 0}\frac{2\sin^2(h/2)}{h^2}}{\lim_{h \to 0}\frac{\sin h}{h}} = \frac{\frac{1}{2} \times \left(\lim_{\frac{h}{2} \to 0}\frac{\sin(h/2)}{h/2}\right)^2}{1} = \frac{1}{2} \times 1^2 = \frac{1}{2} \quad \text{(Ans.)}$$

$10(iv)$ Evaluate: $$\lim_{x \to 0}\frac{1-\cos 4x}{x^2}$$

Solution

$$\lim_{x \to 0}\frac{1-\cos 4x}{x^2} = \lim_{x \to 0}\frac{2\sin^2 2x}{x^2} = 8 \times \lim_{x \to 0}\frac{\sin^2 2x}{(2x)^2}$$

Substituting $z = 2x$, as $x \to 0$, $z \to 0$:

$$= 8 \times \left(\lim_{z \to 0}\frac{\sin z}{z}\right)^2 = 8 \times 1^2 = 8 \quad \text{(Ans.)}$$

$10(v)$ Evaluate: $$\lim_{h \to 0}\frac{\cos ah-\cos bh}{h^2}$$

Solution

Using transformation formula $\cos C - \cos D = 2\sin\left(\frac{C+D}{2}\right)\sin\left(\frac{D-C}{2}\right)$:

$$\lim_{h \to 0}\frac{\cos ah-\cos bh}{h^2} = \lim_{h \to 0}\frac{2\sin \frac{(a+b)h}{2} \sin\frac{(b-a)h}{2}}{h^2}$$ $$= 2 \times \lim_{h \to 0}\frac{\sin \frac{(a+b)h}{2}}{\frac{(a+b)h}{2}} \times \lim_{h \to 0}\frac{\sin \frac{(b-a)h}{2}}{\frac{(b-a)h}{2}} \times \frac{(a+b)(b-a)}{4}$$ $$= 2 \times \left(\lim_{y \to 0} \frac{\sin y}{y}\right) \times \left(\lim_{z \to 0} \frac{\sin z}{z}\right) \times \left(\frac{b^2-a^2}{4}\right)$$ $$= 2 \times 1 \times 1 \times \frac{b^2-a^2}{4} = \frac{b^2-a^2}{2} \quad \text{(Ans.)}$$

Substitution Note: Let $y = \frac{(a+b)h}{2} \implies y \to 0$ as $h \to 0$, and $z = \frac{(b-a)h}{2} \implies z \to 0$ as $h \to 0$.

$10(vi)$ Evaluate: $$\lim_{x \to 0}\frac{\tan^{-1}x}{x}$$

Solution

Let $\tan^{-1}x = \theta \implies x = \tan \theta$. As $x \to 0$, $\theta \to 0$.

$$L = \lim_{\theta \to 0} \left(\frac{\theta}{\tan \theta}\right) = \lim_{\theta \to 0}\left(\frac{\theta}{\sin \theta} \times \cos \theta\right)$$ $$= \frac{1}{\lim_{\theta \to 0}\left(\frac{\sin \theta}{\theta}\right)} \times \lim_{\theta \to 0} \cos \theta = \frac{1}{1} \times \cos 0 = 1 \times 1 = 1 \quad \text{(Ans.)}$$

$10(vii)$ Evaluate: $$\lim_{x \to 0}\frac{x\tan 4x}{1-\cos 4x}$$

Solution

$$\lim_{x \to 0}\frac{x\tan 4x}{1-\cos 4x} = \lim_{x \to 0}\frac{x \sin 4x}{1-\cos 4x} \times \lim_{x \to 0}\frac{1}{\cos 4x}$$ $$= \lim_{x \to 0}\frac{2x\sin 2x\cos 2x}{2\sin^2 2x} \times \frac{1}{\cos 0}$$ $$= \lim_{x \to 0}\frac{x \sin 2x}{\left(\frac{\sin 2x}{2x}\right)^2 \times 4x^2} \times \lim_{x \to 0} \cos 2x \times 1$$ $$= \frac{1}{\left(\lim_{2x \to 0}\frac{\sin 2x}{2x}\right)^2} \times \lim_{2x \to 0}\frac{\sin 2x}{2x} \times \frac{1}{2} \times \cos 0$$ $$= \frac{1}{1} \times 1 \times \frac{1}{2} \times 1 = \frac{1}{2} \quad \text{(Ans.)}$$

$10(viii)$ Evaluate: $$\lim_{x \to 0}\frac{x\tan x}{1-\cos x}$$

Standard Solution

$$\lim_{x \to 0}\frac{x\tan x}{1-\cos x} = \lim_{x \to 0}\frac{x \sin x}{1-\cos x} \times \lim_{x \to 0}\frac{1}{\cos x}$$ $$= \lim_{x \to 0}\frac{2x\sin(x/2)\cos(x/2)}{2\sin^2(x/2)} \times \frac{1}{\cos 0}$$ $$= \lim_{x \to 0}\frac{x}{\sin(x/2)} \times 1 = \frac{1}{\lim_{\frac{x}{2} \to 0}\frac{\sin(x/2)}{x/2} \times \frac{1}{2}} = \frac{1}{1 \times \frac{1}{2}} = 2 \quad \text{(Ans.)}$$

Alternative Rationalization Method

$$\lim_{x \to 0}\frac{x\tan x}{1-\cos x} = \lim_{x \to 0}\frac{x \tan x(1+\cos x)}{(1-\cos x)(1+\cos x)}$$ $$= \lim_{x \to 0}\frac{x\sin x(1+\cos x)}{\cos x(1-\cos^2 x)} = \lim_{x \to 0}\frac{x\sin x(1+\cos x)}{\cos x \times \sin^2 x}$$ $$= \lim_{x \to 0}\frac{x}{\sin x} \times \lim_{x \to 0}\frac{1+\cos x}{\cos x} = 1 \times \frac{1+\cos 0}{\cos 0} = 1 \times \frac{1+1}{1} = 2 \quad \text{(Ans.)}$$