Differentiation (Part-38) | S N De

10(ix)–10(x). Derivative Limits from First Principles

$10(ix)$ Evaluate: $$\lim_{h \to 0}\frac{\sin(x+h)-\sin x}{h}$$

Solution

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

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

$10(x)$ Evaluate: $$\lim_{h \to 0}\frac{\tan(x+h)-\tan x}{h}$$

Solution

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

10(xi)–10(xiv). Advanced Trigonometric Limits at $x \to 0$

$10(xi)$ Evaluate: $$\lim_{x \to 0}\frac{\sin(x^2+4x)}{x^3-5x^2+2x}$$

Solution

$$\lim_{x \to 0}\frac{\sin(x^2+4x)}{x^3-5x^2+2x} = \lim_{x \to 0}\frac{\sin(x^2+4x)}{x^2+4x} \times \lim_{x \to 0}\frac{x^2+4x}{x^3-5x^2+2x}$$ $$= \lim_{z \to 0}\frac{\sin z}{z} \times \lim_{x \to 0} \left(\frac{x(x+4)}{x(x^2-5x+2)}\right)$$ $$= 1 \times \lim_{x \to 0}\frac{x+4}{x^2-5x+2} = \frac{0+4}{0^2-5(0)+2} = \frac{4}{2} = 2 \quad \text{(Ans.)}$$

Substitution Note: Let $z = x^2+4x$. As $x \to 0$, $z \to 0$.

$10(xii)$ Evaluate: $$\lim_{x \to 0}\frac{\sin x(1- \cos x)}{x^3}$$

Solution

$$\lim_{x \to 0}\frac{\sin x(1- \cos x)}{x^3} = \lim_{x \to 0}\frac{\sin x}{x} \times \lim_{x \to 0}\left(\frac{1-\cos x}{x^2}\right)$$ $$= 1 \times \lim_{x \to 0} \frac{2 \sin^2(x/2)}{x^2} = \frac{2}{4} \times \left(\lim_{\frac{x}{2} \to 0}\frac{\sin(x/2)}{x/2}\right)^2$$ $$= \frac{1}{2} \times 1^2 = \frac{1}{2} \quad \text{(Ans.)}$$

$10(xiii)$ Evaluate: $$\lim_{x \to 0}\tan 5x \csc 4x$$

Solution

$$\lim_{x \to 0}\tan 5x \csc 4x = \lim_{x \to 0}\left(\frac{\sin 5x}{\cos 5x} \times \frac{1}{\sin 4x}\right)$$ $$= \lim_{x \to 0}\left(\frac{\sin 5x}{5x} \times 5x\right) \times \lim_{x \to 0}\frac{1}{\cos 5x} \times \frac{1}{\lim_{x \to 0}\left(\frac{\sin 4x}{4x} \times 4x\right)}$$ $$= \lim_{x \to 0}\left(\frac{5x}{4x}\right) \times \lim_{5x \to 0}\frac{\sin 5x}{5x} \times \frac{1}{\cos 0} \times \frac{1}{\lim_{4x \to 0}\frac{\sin 4x}{4x}}$$ $$= \frac{5}{4} \times 1 \times 1 \times 1 = \frac{5}{4} \quad \text{(Ans.)}$$

$10(xiv)$ Evaluate: $$\lim_{x \to 0}\frac{2\sin x-\sin 2x}{x^3}$$

Solution

$$\lim_{x \to 0}\frac{2\sin x-\sin 2x}{x^3} = \lim_{x \to 0}\frac{2\sin x - 2\sin x\cos x}{x^3} = \lim_{x \to 0}\frac{2\sin x(1-\cos x)}{x^3}$$ $$= 2 \times \lim_{x \to 0}\frac{\sin x}{x} \times \lim_{x \to 0}\frac{1-\cos x}{x^2}$$ $$= 2 \times 1 \times \lim_{x \to 0}\frac{2\sin^2(x/2)}{\frac{x^2}{4} \times 4} = 2 \times \frac{2}{4} \times \left(\lim_{\frac{x}{2} \to 0}\frac{\sin(x/2)}{x/2}\right)^2$$ $$= 1 \times 1^2 = 1 \quad \text{(Ans.)}$$

10(xv)–10(xviii). Limits Involving Angle Substitutions & Transformations

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

Solution

$$\lim_{x \to a}\frac{1-\cos(x-a)}{(x-a)^2} = \lim_{x \to a}\frac{2\sin^2\left(\frac{x-a}{2}\right)}{(x-a)^2}$$

Substituting $z = \frac{x-a}{2}$, as $x \to a$, $z \to 0$:

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

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

Solution

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

$10(xvii)$ Evaluate: $$\lim_{x \to 0}\frac{1-\cos 4x}{1-\cos 5x}$$

Solution

$$\lim_{x \to 0}\frac{1-\cos 4x}{1-\cos 5x} = \lim_{x \to 0}\frac{2\sin^2 2x}{2\sin^2\left(\frac{5x}{2}\right)} = \lim_{x \to 0}\frac{\sin^2 2x}{\sin^2\left(\frac{5x}{2}\right)}$$ $$= \lim_{x \to 0}\frac{\left(\frac{\sin 2x}{2x} \times 2x\right)^2}{\left(\frac{\sin(5x/2)}{5x/2} \times \frac{5x}{2}\right)^2} = \lim_{x \to 0}\frac{4x^2}{\frac{25x^2}{4}} \times \frac{\left(\lim_{2x \to 0}\frac{\sin 2x}{2x}\right)^2}{\left(\lim_{\frac{5x}{2} \to 0}\frac{\sin(5x/2)}{5x/2}\right)^2}$$ $$= \frac{16}{25} \times \frac{1^2}{1^2} = \frac{16}{25} \quad \text{(Ans.)}$$

$10(xviii)$ Evaluate: $$\lim_{x \to \pi} \frac{1+\cos x}{\pi-x}$$

Solution

$$\lim_{x \to \pi}\frac{1+\cos x}{\pi-x} = \lim_{x \to \pi}\frac{2\cos^2(x/2)}{\pi-x}$$

Substituting $z = \pi - x \implies x = \pi - z$. As $x \to \pi$, $z \to 0$:

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