# Exponential and Logarithmic Limits

Evaluation of trigonometric limits:

In order to evaluate trigonometric limits the following results are very useful:

1. $$\underset{x\to 0}{\mathop{\lim }}\,$$sinx/x = 1
2. $$\underset{x\to 0}{\mathop{\lim }}\,$$tanx/x = 1
3. $$\underset{x\to 0}{\mathop{\lim }}\,$$sin⁻¹x/x = 1
4. $$\underset{x\to 0}{\mathop{\lim }}\,$$tan⁻¹x/x = 1
5. $$\underset{x\to 0}{\mathop{\lim }}\,$$sinx⁰/x = π/180
6. $$\underset{x\to 0}{\mathop{\lim }}\,$$sin (x – a)/(x – a) = 1
7. $$\underset{x\to 0}{\mathop{\lim }}\,$$tan (x – a)/(x – a) = 1
8. $$\underset{x\to 0}{\mathop{\lim }}\,$$|sinx|/x does not exist
9. $$\underset{x\to 0}{\mathop{\lim }}\,$$|tanx|/x does not exist
10. $$\underset{x\to a}{\mathop{\lim }}\,$$|sin (x – a)|/(x – a) does not exist
11. $$\underset{x\to a}{\mathop{\lim }}\,$$|tan (x – a)|/(x – a) does not exist.

Expansions useful in evaluation of limits:

• (1 + x)ⁿ = 1 + nx + n [(n – 1)/2!] x² + [n (n – 1) (n – 2)/3!] x³ + …
• eˣ = 1 + x/1! + x²/2! + x³/3!
• aˣ = 1 + x/ 1! [logₑᵃ] + x²/2! [x²(logₑᵃ)²] + …
• log (1 + x) = x – x²/2 + x³/3 – x⁴/4 + …
• log (1 – x) = – x – x²/2 – x³/3 – x⁴/4 + …
• sinx = x + x³/3! + x⁵/5! + …
• cos x = 1 – x²/2! + x⁴/4! + …
• tan x = x + x³/3 + 2/15 x⁵ + …
• sin⁻¹ x = x + ½ x³/3 + ½ ¾ x⁵/5 + …
• tan⁻¹ x = x – x³/3 + x⁵/5 + …
• sec⁻¹ x = 1 + x²/2! + 5 x⁴/4! + …

Evaluate: $$\underset{x\to {{\tan }^{-1}}3}{\mathop{\lim }}\,$$ (tan² x – 2 tanx – 3)/ (tan² x – 4 tanx + 3)

Solution: we have,

$$\underset{x\to {{\tan }^{-1}}3}{\mathop{\lim }}\,$$ (tan x – 2 tanx – 3)/ (tan² x – 4 tanx + 3)

= $$\underset{x\to {{\tan }^{-1}}3}{\mathop{\lim }}\,$$ (tanx – 3) (tanx + 1)/ (tanx – 3) (tanx – 1)

= $$\underset{x\to {{\tan }^{-1}}3}{\mathop{\lim }}\,$$ (tanx + 1)/ (tanx – 1) = (3 + 1)/ (3 – 1) = 2

Evaluation of exponential and logarithm limits:

(i) $$\underset{x\to a}{\mathop{\lim }}\,$$ (aˣ – 1)/ x = logₑa, a > 0

(ii) $$\underset{x\to a}{\mathop{\lim }}\,$$ logₑ (1 + x)/x = 1

(iii) $$\underset{x\to a}{\mathop{\lim }}\,$$ (eˣ – 1)/x = logₑ = 1

(iv) $$\underset{x\to a}{\mathop{\lim }}\,$$ logₐ (1 + x)/x = logₐe

Evaluate: $$\underset{x\to 0}{\mathop{\lim }}\,\frac{e-{{\left( 1+x \right)}^{\frac{1}{x}}}}{\tan x}$$,

Solution: We have,

$$\underset{x\to 0}{\mathop{\lim }}\,\frac{e-{{\left( 1+x \right)}^{\frac{1}{x}}}}{\tan x}$$,

$$\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\log }_{a}}\left( 1+x \right)}{x}={{\log }_{a}}e\frac{e-{{e}^{\left( \frac{1}{x} \right).\log \left( 1+x \right)}}}{\tan x}$$,

= $$\underset{x\to 0}{\mathop{\lim }}\,\frac{e-{{e}^{\left( 1-\frac{x}{2}+\frac{{{x}^{2}}}{3}-\frac{{{x}^{3}}}{4}.. \right)}}}{\tan x}$$,

= $$-e\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{e}^{-\frac{x}{2}+\frac{{{x}^{2}}}{3}-\frac{{{x}^{3}}}{4}…}}-1}{\frac{-x}{2}+\frac{{{x}^{2}}}{3}-\frac{{{x}^{3}}}{4}…} \right)\left( \frac{-\frac{1}{2}+\frac{x}{3}-\frac{{{x}^{2}}}{4}..}{\tan x} \right)\,$$,

= $$-e\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{e}^{-\frac{x}{2}+\frac{{{x}^{2}}}{3}-\frac{{{x}^{3}}}{4}..}}-1}{-\frac{x}{2}+\frac{{{x}^{2}}}{3}-\frac{{{x}^{3}}}{4}…} \right)\left( \frac{-\frac{1}{2}+\frac{x}{3}-\frac{{{x}^{2}}}{4}…}{\frac{\tan x}{x}} \right)$$,

= – e x – ½ = e/2