Differentiation (Part-38) | S N De

13. Standard Exponential Limits of the Form $(1+x)^{\frac{1}{x}}$

Standard Identity Given: $\lim_{x \to 0} (1+x)^{\frac{1}{x}} = e$

$13(i)$ Show that: $$\lim_{x \to 0} (1+2x)^{\frac{1}{x}} = e^2$$

Solution

$$\lim_{x \to 0} (1+2x)^{\frac{1}{x}} = \lim_{x \to 0}\left[(1+2x)^{\frac{1}{2x}}\right]^2$$

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

$$= \lim_{z \to 0}\left[(1+z)^{\frac{1}{z}}\right]^2 = \left[\lim_{z \to 0}(1+z)^{\frac{1}{z}}\right]^2 = e^2 \quad \text{(Proved)}$$

$13(ii)$ Show that: $$\lim_{x \to 0} (1-3x)^{\frac{3}{x}} = e^{-9}$$

Solution

$$\lim_{x \to 0} (1-3x)^{\frac{3}{x}} = \lim_{x \to 0}\left[(1-3x)^{\frac{1}{-3x}}\right]^{-9}$$

Substituting $z = -3x$, as $x \to 0$, $z \to 0$:

$$= \left[\lim_{z \to 0}(1+z)^{\frac{1}{z}}\right]^{-9} = e^{-9} \quad \text{(Proved)}$$

$13(iii)$ Show that: $$\lim_{x \to 1} x^{\frac{1}{1-x}} = e^{-1}$$

Solution

Substituting $z = x - 1 \implies x = 1 + z$. As $x \to 1$, $z \to 0$:

$$\lim_{x \to 1} x^{\frac{1}{1-x}} = \lim_{z \to 0} (1+z)^{-\frac{1}{z}}$$ $$= \left[\lim_{z \to 0} (1+z)^{\frac{1}{z}}\right]^{-1} = e^{-1} \quad \text{(Proved)}$$

$13(iv)$ Show that: $$\lim_{x \to 0} (1+4x)^{\frac{x+2}{x}} = e^8$$

Solution

$$\lim_{x \to 0} (1+4x)^{\frac{x+2}{x}} = \lim_{x \to 0}(1+4x)^{1+\frac{2}{x}} = \lim_{x \to 0} \left[(1+4x) \times (1+4x)^{\frac{2}{x}}\right]$$ $$= \lim_{x \to 0}(1+4x) \cdot \lim_{x \to 0}(1+4x)^{\frac{2}{x}}$$ $$= 1 \cdot \lim_{x \to 0} \left[(1+4x)^{\frac{1}{4x}}\right]^8$$

Substituting $z = 4x$, as $x \to 0$, $z \to 0$:

$$= \left[\lim_{z \to 0} (1+z)^{\frac{1}{z}}\right]^8 = e^8 \quad \text{(Proved)}$$

14. Limits Involving Piecewise Functions

$14.$ If $f(x) = \lim_{n \to \infty}\frac{1}{1+n\sin^2 \pi x}$, find $f(x)$ for all real values of $x$.

Solution

We analyze $f(x)$ under two cases depending on whether $x$ is an integer or non-integer:

Case 1: When $x$ is an integer ($x \in \mathbb{Z}$)

If $x$ is any integer, then $\sin(\pi x) = 0$.

$$\therefore f(x) = \lim_{n \to \infty}\frac{1}{1 + n \cdot 0} = \frac{1}{1} = 1$$

Case 2: When $x$ is non-integral / fractional ($x \notin \mathbb{Z}$)

For any fractional value of $x$, $\sin(\pi x) \neq 0 \implies \sin^2(\pi x) > 0$.

As $n \to \infty$, $(1 + n\sin^2 \pi x) \to \infty$.

$$\therefore f(x) = \lim_{n \to \infty}\frac{1}{1+n\sin^2 \pi x} = 0$$

Final Piecewise Expression:

$$f(x) = \begin{cases} 1, & \text{if } x \in \mathbb{Z} \text{ (integer)} \\ 0, & \text{if } x \notin \mathbb{Z} \text{ (non-integer)} \end{cases}$$

15–16. Advanced Limits and L'Hospital's Rule

$15.$ Evaluate: $$\lim_{x \to \frac{\pi}{4}}\frac{4\sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2x}$$

Solution

$$\lim_{x \to \frac{\pi}{4}}\frac{4\sqrt{2}-(\cos x+\sin x)^5}{1-\sin 2x} = \lim_{x \to \frac{\pi}{4}}\frac{4\sqrt{2}-\left\{(\cos x+\sin x)^2\right\}^{\frac{5}{2}}}{1-\sin 2x}$$

Since $(\cos x + \sin x)^2 = \cos^2 x + \sin^2 x + 2\sin x\cos x = 1 + \sin 2x$:

$$= \lim_{x \to \frac{\pi}{4}}\frac{(2)^{\frac{5}{2}}-(1+\sin 2x)^{\frac{5}{2}}}{2-(1+\sin 2x)}$$

Substituting $z = 1 + \sin 2x$. As $x \to \frac{\pi}{4}$, $z \to 2$:

$$= \lim_{z \to 2}\frac{2^{5/2}-z^{5/2}}{2-z} = \lim_{z \to 2}\frac{z^{5/2}-2^{5/2}}{z-2}$$

Using standard algebraic limit formula $\lim_{z \to a} \frac{z^n - a^n}{z - a} = n a^{n-1}$:

$$= \frac{5}{2} \cdot 2^{\frac{5}{2}-1} = \frac{5}{2} \cdot 2^{\frac{3}{2}} = 5 \sqrt{2} \quad \text{(Ans.)}$$

Substitution Note: Let $z = 1 + \sin 2x$. As $x \to \frac{\pi}{4}$, $\sin 2x \to 1 \implies z \to 2$.

$16.$ Evaluate: $$\lim_{x \to 4} \frac{(\cos \alpha)^x-(\sin\alpha)^x-\cos 2\alpha}{x-4} \quad \left(0 < \alpha < \frac{\pi}{4}\right)$$

Solution

Let $L = \lim_{x \to 4} \frac{(\cos \alpha)^x-(\sin\alpha)^x-\cos 2\alpha}{x-4} \quad \dots (1)$

At $x = 4$, the numerator gives $\cos^4 \alpha - \sin^4 \alpha - \cos 2\alpha = (\cos^2 \alpha - \sin^2 \alpha)(\cos^2 \alpha + \sin^2 \alpha) - \cos 2\alpha = \cos 2\alpha - \cos 2\alpha = 0$.

Since $L$ is in $\left[\frac{0}{0}\right]$ indeterminate form, applying L'Hospital's Rule by differentiating numerator and denominator with respect to $x$:

$$L = \lim_{x \to 4}\frac{\frac{d}{dx}\left[(\cos\alpha)^x - (\sin\alpha)^x - \cos 2\alpha\right]}{\frac{d}{dx}(x-4)}$$ $$= \lim_{x \to 4}\frac{\cos^x \alpha \cdot \ln(\cos\alpha) - \sin^x \alpha \cdot \ln(\sin\alpha)}{1}$$

Substituting $x = 4$ into the expression:

$$= \cos^4 \alpha \cdot \ln(\cos\alpha) - \sin^4 \alpha \cdot \ln(\sin\alpha) \quad \text{(Ans.)}$$