Differentiation (Part-38) | S N De

8. Bounded Function Limits and Trigonometric Theorems

$8(i)$ Prove: $$\lim_{x \to 0} x\cos \frac{1}{x} = 0$$

Solution

We know that the cosine function is bounded between $-1$ and $1$:

$$-1 \le \cos \frac{1}{x} \le 1$$

Also, we have $\lim_{x \to 0} x = 0$.

By the Squeeze Theorem (or the product of a bounded function and a infinitesimal sequence), the limit of a bounded function multiplied by a function tending to $0$ is equal to $0$.

$$\therefore \lim_{x \to 0} x\cos \frac{1}{x} = 0 \quad \text{(Proved)}$$

$8(ii)$ Prove: $$\lim_{x \to 0} x^2\sin \frac{1}{x} = 0$$

Solution

We know that the sine function is bounded between $-1$ and $1$:

$$-1 \le \sin \frac{1}{x} \le 1$$

Also, we have $\lim_{x \to 0} x^2 = 0$.

Therefore, the limit of a bounded function multiplied by a function that approaches $0$ must be $0$.

$$\therefore \lim_{x \to 0} x^2\sin \frac{1}{x} = 0 \quad \text{(Proved)}$$

$8(iii)$ Prove: $$\lim_{x \to 0}\frac{x^2\sin\frac{1}{x}}{\sin x} = 0$$

Solution

$$\lim_{x \to 0}\frac{x^2\sin\frac{1}{x}}{\sin x} = \lim_{x \to 0}\frac{x \sin\frac{1}{x}}{\frac{\sin x}{x}}$$

Using the result from part $8(i)$ where $\lim_{x \to 0} x\sin\frac{1}{x} = 0$ and the standard limit $\lim_{x \to 0}\frac{\sin x}{x} = 1$:

$$= \frac{0}{1} = 0 \quad \text{(Proved)}$$

$8(iv)$ Prove: $$\lim_{x \to \infty}\frac{\sin x}{x} = 0$$

Solution

Let $L = \lim_{x \to \infty}\frac{\sin x}{x}$. Substituting $x = \frac{1}{z}$, as $x \to \infty$, $z \to 0$:

$$L = \lim_{z \to 0} z\sin\frac{1}{z} \quad \dots (1)$$

Since the sine function is bounded ($-1 \le \sin\frac{1}{z} \le 1$) and $\lim_{z \to 0} z = 0$,

$$\lim_{z \to 0} z\sin \frac{1}{z} = 0$$

Hence, from equation $(1)$, $L = 0 \quad \text{(Proved)}$.

$8(v)$ Prove: $$\lim_{x \to \infty}\frac{\cos x}{x} = 0$$

Solution

Let $L = \lim_{x \to \infty}\frac{\cos x}{x}$. Substituting $z = \frac{1}{x}$, as $x \to \infty$, $z \to 0$:

$$L = \lim_{z \to 0} z \cos \frac{1}{z} \quad \dots (1)$$

Using the result established in part $8(i)$, equation $(1)$ yields:

$$L = 0 \quad \text{(Proved)}$$

9–10. Trigonometric Limits using Variable Substitution

$9.$ By substituting $z = x - \frac{\pi}{4}$, show that: $$\lim_{x \to \pi/4}\frac{1-\tan x}{1-\sqrt{2} \sin x} = 2$$

Solution

Let $z = x - \frac{\pi}{4} \implies x = z + \frac{\pi}{4}$. As $x \to \frac{\pi}{4}$, $z \to 0$.

$$\lim_{x \to \pi/4}\frac{1-\tan x}{1-\sqrt{2} \sin x} = \lim_{z \to 0}\frac{1-\tan\left(z+\frac{\pi}{4}\right)}{1-\sqrt{2} \sin\left(z+\frac{\pi}{4}\right)}$$

Simplifying the trigonometric expressions in the numerator and denominator:

$$= \lim_{z \to 0}\frac{\frac{-2\tan z}{1-\tan z}}{1-\cos z-\sin z}$$ $$= \lim_{z \to 0}\frac{-2\tan z}{(1-\tan z)\left(2\sin^2\frac{z}{2}-2\sin\frac{z}{2}\cos \frac{z}{2}\right)}$$ $$= \lim_{z \to 0} \frac{-2\frac{\sin z}{\cos z}}{(1-\tan z) \cdot 2\sin \frac{z}{2}\left(\sin\frac{z}{2}-\cos\frac{z}{2}\right)}$$ $$= \lim_{z \to 0}\frac{-\sin z}{\cos z(1-\tan z)\sin \frac{z}{2}\left(\sin\frac{z}{2}-\cos\frac{z}{2}\right)}$$ $$= \lim_{z \to 0}\frac{-2\sin \frac{z}{2} \cos\frac{z}{2}}{\cos z (1-\tan z)\sin \frac{z}{2}\left(\sin\frac{z}{2}-\cos\frac{z}{2}\right)}$$ $$= \lim_{z \to 0}\frac{-2\cos \frac{z}{2}}{\cos z(1-\tan z)\left(\sin\frac{z}{2}-\cos\frac{z}{2}\right)}$$ $$= \frac{-2 \times 1}{1(1-0)(0-1)} = 2 \quad \text{(Showed)}$$

Trigonometric Simplifications Used:
1. $1-\tan\left(z+\frac{\pi}{4}\right) = 1-\frac{\tan z+1}{1-\tan z} = \frac{-2\tan z}{1-\tan z}$
2. $1-\sqrt{2}\sin\left(z+\frac{\pi}{4}\right) = 1-\sqrt{2}\left(\frac{\sin z + \cos z}{\sqrt{2}}\right) = 1-\sin z-\cos z$

$10.$ By substituting $x = \theta - \frac{\pi}{4}$, prove that: $$\lim_{\theta \to \pi/4}\frac{\sin \theta-\cos\theta}{\theta-\pi/4} = \sqrt{2}$$

Solution

$$\lim_{\theta \to \pi/4}\frac{\sin \theta-\cos\theta}{\theta-\pi/4} = \lim_{\theta \to \pi/4} \sqrt{2} \cdot \left(\frac{\frac{1}{\sqrt{2}}\sin\theta-\frac{1}{\sqrt{2}}\cos \theta}{\theta-\pi/4}\right)$$ $$= \sqrt{2} \cdot \lim_{\theta \to \pi/4}\left[\frac{\sin\left(\theta-\frac{\pi}{4}\right)}{\theta-\frac{\pi}{4}}\right]$$

Substituting $x = \theta - \frac{\pi}{4}$, as $\theta \to \frac{\pi}{4}$, $x \to 0$:

$$= \sqrt{2} \cdot \lim_{x \to 0} \frac{\sin x}{x} = \sqrt{2} \times 1 = \sqrt{2} \quad \text{(Proved)}$$

11. Determination of Unknown Constants

$11.$ If $\lim_{x \to 2}\frac{ax^2-b}{x-2} = 4$, find the values of $a$ and $b$.

Solution

Since the limit exists and the denominator vanishes at $x = 2$, $x = 2$ must also be a root of the numerator $(ax^2 - b)$.

Let $(ax^2 - b) = (x - 2)(cx + d) \quad \dots (1)$

Expanding and comparing coefficients on both sides of $(1)$:

$$a = c, \quad d = 2c, \quad b = 2d \quad \dots (2)$$

Now, evaluating the limit:

$$\lim_{x \to 2}\frac{ax^2-b}{x-2} = 4 \implies \lim_{x \to 2}\frac{(x-2)(cx+d)}{x-2} = 4$$ $$\implies 2c + d = 4$$

Substituting $d = 2c$ into $2c + d = 4$ gives $4c = 4 \implies c = 1, d = 2 \quad \dots (3)$.

Using $(2)$ and $(3)$:

$$a = 1, \quad b = 2 \times 2 = 4 \quad \text{(Ans.)}$$

12. Exponential and Logarithmic Standard Limits

$12(i)$ Prove that: $$\lim_{x \to 1}\frac{4^x-4}{x-1} = 8 \log_e 2$$

Solution

$$\lim_{x \to 1}\frac{4^x-4}{x-1} = \lim_{x \to 1}\frac{4(4^{x-1}-1)}{x-1}$$

Substituting $z = x - 1$, as $x \to 1$, $z \to 0$:

$$= 4 \times \lim_{z \to 0}\frac{4^z-1}{z}$$ $$= 4 \times \log_e 4 = 4 \times \log_e 2^2 = 4 \times 2\log_e 2 = 8\log_e 2 \quad \text{(Proved)}$$

$12(ii)$ Prove that: $$\lim_{x \to 2} \frac{\log(2x-3)}{2(x-2)} = 1$$

Solution

Substituting $z = x - 2 \implies x = z + 2$. As $x \to 2$, $z \to 0$:

$$\lim_{x \to 2} \frac{\log(2x-3)}{2(x-2)} = \lim_{z \to 0} \frac{\log[2(z+2)-3]}{2z}$$ $$= \lim_{z \to 0} \frac{\log(2z+1)}{2z}$$

Letting $p = 2z$, as $z \to 0$, $p \to 0$:

$$= \lim_{p \to 0} \frac{\log(p+1)}{p} = 1 \quad \text{(Proved)}$$