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Q20:
Evaluate : $lim_{x->\frac{\pi}{3}} \frac{\sqrt{1-cos(6x)}}{\sqrt{2}(\frac{\pi}{3}-x)}$
Solution :
Given Variables & Initial Setup
We are tasked to evaluate the limit: $$L = \lim_{x \to \frac{\pi}{3}} \frac{\sqrt{1 - \cos(6x)}}{\sqrt{2} \left( \frac{\pi}{3} - x \right)}$$
Step 1: Trigonometric Simplification
Utilize the half-angle identity: $1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right)$. Applying this to the term $\cos(6x)$:
$$1 - \cos(6x) = 2\sin^2\left(\frac{6x}{2}\right) = 2\sin^2(3x)$$[By trigonometric identity: $1 - \cos(2\theta) = 2\sin^2(\theta)$]
Step 2: Substitution into the Expression
Substitute the simplified numerator into the original expression:
$$L = \lim_{x \to \frac{\pi}{3}} \frac{\sqrt{2\sin^2(3x)}}{\sqrt{2} \left( \frac{\pi}{3} - x \right)} = \lim_{x \to \frac{\pi}{3}} \frac{\sqrt{2} \cdot |\sin(3x)|}{\sqrt{2} \left( \frac{\pi}{3} - x \right)} = \lim_{x \to \frac{\pi}{3}} \frac{|\sin(3x)|}{\frac{\pi}{3} - x}$$Step 3: Change of Variable
Let $h = \frac{\pi}{3} - x$. As $x \to \frac{\pi}{3}$, $h \to 0$. Rearranging gives $x = \frac{\pi}{3} - h$. Substitute $x$ into the sine function:
$$\sin(3x) = \sin(3(\frac{\pi}{3} - h)) = \sin(\pi - 3h)$$[Per the supplementary angle property: $\sin(\pi - \theta) = \sin(\theta)$]
$$\sin(\pi - 3h) = \sin(3h)$$Step 4: Evaluating the Limit
Since we are evaluating the limit as $x \to \frac{\pi}{3}$, $3x$ approaches $\pi$. The function $\sin(3x)$ is positive for values slightly less than $\pi$ and negative for values slightly greater than $\pi$. Thus, we must evaluate the Left-Hand Limit (LHL) and Right-Hand Limit (RHL).
Case 1: $x \to \frac{\pi}{3}^-$ (which implies $h \to 0^+$)
$$L_{LHL} = \lim_{h \to 0^+} \frac{\sin(3h)}{h} = 3 \cdot \lim_{3h \to 0} \frac{\sin(3h)}{3h} = 3(1) = 3$$[By Fundamental Limit Theorem: $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$]
Case 2: $x \to \frac{\pi}{3}^+$ (which implies $h \to 0^-$)
As $x > \frac{\pi}{3}$, $\sin(3x)$ becomes negative, so $|\sin(3x)| = -\sin(3x)$.
$$L_{RHL} = \lim_{h \to 0^-} \frac{-\sin(3h)}{h} = -3 \cdot \lim_{3h \to 0} \frac{\sin(3h)}{3h} = -3(1) = -3$$Step 5: Conclusion on Existence
Since the Left-Hand Limit ($3$) does not equal the Right-Hand Limit ($-3$), the limit does not exist in the general sense.
Final Solution: The limit does not exist because the Left-Hand Limit is 3 and the Right-Hand Limit is -3.
More Questions from Class 11 Mathematics Limits and Derivatives EXERCISE 13.3
- Q1: Evaluate : $lim_{x->3} \frac{x^2 - 9}{x - 3}$
- Q2: Evaluate : $lim_{x->\frac{1}{2}} \frac{4x^2 - 1}{2x - 1}$
- Q3: Evaluate : $lim_{h->0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$
- Q4: Evaluate : $lim_{x->0} \frac{\sqrt[3]{(x+2)} - \sqrt[3]{2}}{x}$
- Q5: Evaluate : $lim_{x->0} \frac{\sqrt[6]{(1+x)} - 1}{\sqrt[21]{(1+x)} - 1}$
- Q6: Evaluate : $lim_{x->a} \frac{(x+2)^5 - (a+2)^5}{x-a}$
- Q7: Evaluate : $lim_{x->1} \frac{x^4 - \sqrt{x}}{x - 1}$
- Q8: Evaluate : $lim_{x->2} \frac{x^2 - 4}{\sqrt{3x-2} - \sqrt{x+2}}$
- Q9: Evaluate : $lim_{x->\sqrt{2}} \frac{x^4 - 4}{x^2 + 3\sqrt{2}x - 8}$
- Q10: Evaluate : $lim_{x->1} \frac{x^7 - 2x^5 + 1}{x^3 - 3x^2 + 2}$
- Q11: Evaluate : $lim_{x->0} \frac{\sqrt[3]{1+x} - \sqrt[3]{1-x}}{x}$
- Q12: Evaluate : $lim_{x->-3} \frac{x^3 + 27}{x^5 + 243}$
- Q13: Evaluate : $lim_{x->\frac{1}{2}} (\frac{8x-3}{2x-1} - \frac{4x^2+1}{4x^2-1})$
- Q14: Find 'n', if $lim_{x->2} \frac{x^n - 2^n}{x-2} = 80, n \in N$
- Q15: Evaluate : $lim_{x->a} \frac{sin(3x)}{sin(7x)}$
- Q16: Evaluate : $lim_{x->0} \frac{sin^2(2x)}{sin^2(4x)}$
- Q17: Evaluate : $lim_{x->0} \frac{1-cos(2x)}{x^2}$
- Q18: Evaluate : $lim_{x->0} \frac{2sin(x) - sin(2x)}{x^3}$
- Q19: Evaluate : $lim_{x->0} \frac{1-cos(mx)}{1-cos(nx)}$
- Q21: Evaluate : $lim_{x->\frac{\pi}{4}} \frac{sin(x) - cos(x)}{x - \frac{\pi}{4}}$
- Q22: Evaluate : $lim_{x->\frac{\pi}{6}} \frac{\sqrt{3}sin(x) - cos(x)}{x - \frac{\pi}{6}}$
- Q23: Evaluate : $lim_{x->0} \frac{sin(2x) + 3x}{2x + tan(3x)}$
- Q24: Evaluate : $lim_{x->a} \frac{sin(x) - sin(a)}{x-a}$
- Q25: Evaluate : $lim_{x->\frac{\pi}{6}} \frac{cot^2(x) - 3}{cosec(x) - 2}$
- Q26: Evaluate : $lim_{x->0} \frac{\sqrt{2} - \sqrt{1+cos(x)}}{sin^2(x)}$
- Q27: Evaluate : $lim_{x->0} \frac{sin(x) - 2sin(3x) + sin(5x)}{x}$
- Q28: If $lim_{x->1} \frac{x^4-1}{x-1} = lim_{x->k} \frac{x^3 - k^3}{x^2 - k^2}$, then find the value of k.
- Q29: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ \frac{x^4 + x^3 + x^2 + 1}{x} $
- Q30: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ (x + \frac{1}{x})^3 $
- Q31: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ (3x + 5) (1 + tan(x)) $
- Q32: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ (sec(x) – 1) (sec(x) + 1) $
- Q33: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ \frac{3x+4}{5x^2 - 7x + 9} $
- Q34: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ \frac{x^5 - cos(x)}{sin(x)} $
- Q35: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ \frac{x^2 cos(\frac{\pi}{4})}{sin(x)} $
- Q36: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ (ax^2 + cot(x)) (p + q cos(x)) $
- Q37: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ \frac{a+b sin(x)}{c+d cos(x)} $
- Q38: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ (sin(x) + cos(x))^2 $
- Q39: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ (2x – 7)^2 (3x + 5)^3 $
- Q40: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ x^2 sin(x) + cos(2x) $
- Q41: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ sin^3(x) cos^3(x) $
- Q42: Differentiate each of the functions w. r. to x in Exercises 29 to 42. \n $ \frac{1}{ax^2 + bx + c} $
- Q43: Differentiate each of the functions with respect to ‘x’ in Exercises 43 to 46 using first principle. \n $ cos(x^2 + 1) $
- Q44: Differentiate each of the functions with respect to ‘x’ in Exercises 43 to 46 using first principle. \n $ \frac{ax+b}{cx+d} $
- Q45: Differentiate each of the functions with respect to ‘x’ in Exercises 43 to 46 using first principle. \n $ \frac{x^2}{3} $
- Q46: Differentiate each of the functions with respect to ‘x’ in Exercises 43 to 46 using first principle. \n $ x cos(x) $
- Q47: Evaluate each of the following limits in Exercises 47 to 53. \n $lim_{y->0} \frac{(x+y)sec(x+y) - x sec(x)}{y}$
- Q48: Evaluate each of the following limits in Exercises 47 to 53. \n $lim_{x->0} \frac{(sin(\alpha+\beta x) + sin(\alpha-\beta x) - 2sin(\alpha))}{x \cdot (cos(2\beta x) - cos(2\alpha))}$
- Q49: Evaluate each of the following limits in Exercises 47 to 53. \n $lim_{x->\frac{\pi}{4}} \frac{tan^3(x) - tan(\frac{\pi}{4})}{x - \frac{\pi}{4}}$
- Q50: Evaluate each of the following limits in Exercises 47 to 53. \n $lim_{x->\pi} \frac{1-sin(\frac{x}{2})}{cos(\frac{x}{2})(cos(\frac{x}{4}) - sin(\frac{x}{4}))}$
- Q51: Show that $lim_{x->4} \frac{|x-4|}{x-4}$ does not exists
- Q52: Let $ f(x) = \begin{cases} \frac{k \cos(x)}{\pi - 2x} & \text{when } x \neq \frac{\pi}{2} \\ 3 & \text{when } x = \frac{\pi}{2} \end{cases} $ and if $lim_{x->\frac{\pi}{2}} f(x) = f(\frac{\pi}{2})$, find the value of k.
- Q53: Let $ f(x) = \begin{cases} x+2 & x \leq -1 \\ cx^2 & x > -1 \end{cases} $, find ‘c’ if $lim_{x->-1} f(x)$ exists.
- Q54: $lim_{x->\pi} \frac{sin(x)}{x-\pi}$ is
- Q55: $lim_{x->0} \frac{cos(2x)-1}{cos(x)-1}$ is
- Q56: $lim_{x->0} \frac{(1+x)^n - 1}{x}$ is
- Q57: $lim_{x->1} \frac{x^m-1}{x^n-1}$ is
- Q58: $lim_{x->0} \frac{1-cos(4\theta)}{1-cos(6\theta)}$ is
- Q59: $lim_{x->0} \frac{cosec(x) - cot(x)}{x}$ is
- Q60: $lim_{x->0} \frac{sin(x)}{\sqrt{x+1}-\sqrt{1-x}}$ is
- Q61: $lim_{x->\frac{\pi}{4}} \frac{sec^2(x) - 2}{tan(x) - 1}$ is
- Q62: $lim_{x->1} \frac{(x+x^2+x^3) - 3}{x-1}$ is
- Q63: If $f(x) = \begin{cases} \frac{sin[x]}{[x]} & [x] \neq 0 \\ 0 & [x]=0 \end{cases}$, where [.] denotes the greatest integer function, then $lim_{x->0} f(x)$ is equal to
- Q64: $lim_{x->0} \frac{|sin(x)|}{x}$ is
- Q65: Let $ f(x) = \begin{cases} x^2-1, & 0 < x < 2 \\ 2x+3, & 2 \leq x < 3 \end{cases} $, the quadratic equation whose roots are $lim_{x->-2} f(x)$ and $lim_{x->2^+} f(x)$ is
- Q66: $lim_{x->0} \frac{tan(x) - x}{x^2 tan(x)}$ is
- Q67: Let $f(x) = x – [x]$; $x \in R$, then $f'(\frac{1}{2})$ is
- Q68: If $y = \sqrt{x} + \frac{1}{\sqrt{x}}$, then $\frac{dy}{dx}$ at x = 1 is
- Q69: If $f(x) = \frac{x-4}{2\sqrt{x}}$, then $f'(1)$ is
- Q70: If $y = \frac{1-x^2/2}{1+x^2/2}$, then $\frac{dy}{dx}$ is
- Q71: If $y = \frac{sin(x) + cos(x)}{sin(x) - cos(x)}$, then $\frac{dy}{dx}$ at x = 0 is
- Q72: If $y = \frac{sin(x+9)}{cos(x)}$, then $\frac{dy}{dx}$ at x = 0 is
- Q73: If $f(x) = 1 + x + \frac{x^2}{2} + ... + \frac{x^{100}}{100}$, then $f'(1)$ is equal to
- Q74: If $f(x) = \frac{x^n - a^n}{x-a}$ for some constant ‘a’, then $f'(a)$ is
- Q75: If $f(x) = x^{100} + x^{99} + ... + x + 1$, then $f'(1)$ is equal to
- Q76: If $f(x) = 1 – x + x^2 – x^3 ... – x^{99} + x^{100}$, then $f'(1)$ is equal to
- Q77: Fill in the blanks in Exercises 77 to 80. \n If $f(x) = \frac{tan(x)}{x-\pi}$, then $lim_{x->\pi} f(x)$ = ______________
- Q78: Fill in the blanks in Exercises 77 to 80. \n $lim_{x->0} \frac{sin(mx)}{tan^3(2x)} = 2$, then m = ______________
- Q79: Fill in the blanks in Exercises 77 to 80. \n if $y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$, then $\frac{dy}{dx}$ = ______________
- Q80: Fill in the blanks
CBSE Solutions for Class 11 Mathematics Limits and Derivatives
Chapters in CBSE - Class 11 Mathematics
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