Evaluation of trigonometric limits:
In order to evaluate trigonometric limits the following results are very useful:
- \(\underset{x\to 0}{\mathop{\lim }}\,\)sinx/x = 1
- \(\underset{x\to 0}{\mathop{\lim }}\,\)tanx/x = 1
- \(\underset{x\to 0}{\mathop{\lim }}\,\)sin⁻¹x/x = 1
- \(\underset{x\to 0}{\mathop{\lim }}\,\)tan⁻¹x/x = 1
- \(\underset{x\to 0}{\mathop{\lim }}\,\)sinx⁰/x = π/180
- \(\underset{x\to 0}{\mathop{\lim }}\,\)sin (x – a)/(x – a) = 1
- \(\underset{x\to 0}{\mathop{\lim }}\,\)tan (x – a)/(x – a) = 1
- \(\underset{x\to 0}{\mathop{\lim }}\,\)|sinx|/x does not exist
- \(\underset{x\to 0}{\mathop{\lim }}\,\)|tanx|/x does not exist
- \(\underset{x\to a}{\mathop{\lim }}\,\)|sin (x – a)|/(x – a) does not exist
- \(\underset{x\to a}{\mathop{\lim }}\,\)|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