CBSE Class 11 Mathematics Limits and Derivatives Assignment Set 09

Question 1. If \(y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty\). Find \(\frac{dy}{dx}\)
Answer: We have \(y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty\) ---------- (i)
Differentiate both sides with respect to \(x\):
\(\frac{dy}{dx} = 0 + \frac{1}{1!} + \frac{2x}{2!} + \frac{3x^2}{3!} + \dots \infty\)
\(\frac{dy}{dx} = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots \infty\)
\(\frac{dy}{dx} = y\) (From eq. (i))
In simple words: Differentiating the infinite series term by term yields the exact same exponential series, which equals \(y\).

Exam Tip: Remember that the series expansion of \(e^x\) is \(1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots\), whose derivative is itself.

Question 2. Evaluate \(\lim_{x \to 3^+} \left( \frac{x}{[x]} \right)\)
Answer: Put \(x = 3 + h\) and \(h \to 0\)
\(= \lim_{h \to 0} \left( \frac{3 + h}{[3 + h]} \right)\)
Since \(3 + h\) is slightly greater than \(3\), its greatest integer value is \(3\) (e.g., \([3.1] = 3\), \([3.01] = 3\)):
\(= \lim_{h \to 0} \left( \frac{3 + h}{3} \right)\)
\(= \frac{3}{3} = 1\)
In simple words: For values just above 3, the greatest integer function evaluates to 3, simplifying the fraction to 1.

Exam Tip: Always evaluate the limit of the greatest integer function from the right side by substituting \(x = c + h\) and applying the property \([c + h] = c\) for small positive \(h\).

Question 3. If \(\lim_{x \to 0} \left( \sin(mx) \cot\left(\frac{x}{\sqrt{3}}\right) \right) = 2\). Find \(m\)
Answer: We have \(\lim_{x \to 0} \left( \sin(mx) \cot\left(\frac{x}{\sqrt{3}}\right) \right) = 2\)
\(= \lim_{x \to 0} \left( \frac{\sin(mx)}{\tan\left(\frac{x}{\sqrt{3}}\right)} \right) = 2\)
\(= \lim_{x \to 0} \left( \frac{\sin(mx)}{mx} \times mx \times \frac{\frac{x}{\sqrt{3}}}{\tan\left(\frac{x}{\sqrt{3}}\right)} \times \frac{1}{\frac{x}{\sqrt{3}}} \right) = 2\)
\(= \frac{1 \times m}{1 \times \frac{1}{\sqrt{3}}} = 2\)
\(\sqrt{3}m = 2\)
\(m = \frac{2}{\sqrt{3}}\)
In simple words: Convert cotangent to tangent, then use standard limit formulas like \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\).

Exam Tip: Adjust the angles in the numerator and denominator to match standard trigonometric limits by multiplying and dividing by the necessary constants.

Question 4. If \(f(x) = 1 - x + x^2 - x^3 \dots - x^{99} + x^{100}\)
Answer: We have \(f(x) = 1 - x + x^2 - x^3 \dots - x^{99} + x^{100}\)
Differentiate both sides with respect to \(x\):
\(f'(x) = 0 - 1 + 2x - 3x^2 \dots - 99x^{98} + 100x^{99}\)
\(f'(1) = -1 + 2 - 3 \dots - 99 + 100\)
\(f'(1) = -(1 + 3 + 5 \dots 99) + (2 + 4 + 6 \dots 100)\)
For the first A.P.: \(a = 1, d = 2, n = 50\)
For the second A.P.: \(a = 2, d = 2, n = 50\)
\(= -\frac{50}{2} [2 + (49 \times 2)] + \frac{50}{2} [4 + (49 \times 2)]\)
\(= -25(100) + 25(102)\)
\(= -2500 + 2550\)
\(f'(1) = 50\)
In simple words: Group the negative and positive terms separately into arithmetic progressions to compute their sums easily.

Exam Tip: When evaluating alternating series derivatives at \(x = 1\), separate positive and negative terms to form standard arithmetic progressions.

Question 5. Let \(f(x) = \begin{cases} x^2 - 1 & : x < 2 \\ 2x + 3 & : 2 \le x < 3 \end{cases}\) find the quadratic curve whose roots are \(\lim_{x \to 2^-} f(x)\) and \(\lim_{x \to 2^+} f(x)\)
Answer: First, find the left-hand limit:
\(\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x^2 - 1)\)
Put \(x = 2 - h\) and \(h \to 0\)
\(= \lim_{h \to 0} ((2 - h)^2 - 1) = 4 - 1 = 3\)
Now, find the right-hand limit:
\(\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x + 3)\)
Put \(x = 2 + h\) and \(h \to 0\)
\(= \lim_{h \to 0} (2(2 + h) + 3) = 4 + 3 = 7\)
Given 3 and 7 are the roots of the quadratic equation:
\(x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0\)
\(x^2 - (3 + 7)x + (3 \times 7) = 0\)
\(x^2 - 10x + 21 = 0\)
In simple words: Evaluate the left and right limits at 2 to get the roots 3 and 7, then construct the quadratic equation using \(x^2 - (\text{sum})x + \text{product} = 0\).

Exam Tip: A quadratic equation with roots \(\alpha\) and \(\beta\) is always given by \(x^2 - (\alpha + \beta)x + \alpha\beta = 0\).

Question 6. Evaluate \(\lim_{x \to \frac{\pi}{3}} \left( \frac{\sqrt{1 - \cos(6x)}}{\sqrt{2}\left(\frac{\pi}{3} - x\right)} \right)\)
Answer: We have \(\lim_{x \to \frac{\pi}{3}} \left( \frac{\sqrt{1 - \cos(6x)}}{\sqrt{2}\left(\frac{\pi}{3} - x\right)} \right)\)
\(= \lim_{x \to \frac{\pi}{3}} \left( \frac{\sqrt{2 - \sin^2(3x)}}{\sqrt{2}\left(\frac{\pi}{3} - x\right)} \right)\)
\(= \lim_{x \to \frac{\pi}{3}} \left( \frac{\sqrt{2}\sin(3x)}{\sqrt{2}\left(\frac{\pi}{3} - x\right)} \right)\)
Put \(x = \frac{\pi}{3} + h\) and \(h \to 0\)
\(= \lim_{h \to 0} \left( \frac{\sin\left(3\left(\frac{\pi}{3} + h\right)\right)}{\frac{\pi}{3} + h - \frac{\pi}{3}} \right)\)
\(= \lim_{h \to 0} \left( \frac{\sin(\pi + 3h)}{h} \right)\)
\(= \lim_{h \to 0} \left( \frac{-\sin(3h)}{h} \right)\)
\(= \lim_{h \to 0} \left( \frac{\sin(3h)}{3h} \right) \times 3\)
\(= 1 \times 3 = 3\)
In simple words: Simplify the trigonometric expression using double angle formulas and substitute \(x = \frac{\pi}{3} + h\) to evaluate the limit as \(h \to 0\).

Exam Tip: Use trigonometric identities to simplify radical expressions before applying standard substitution for limits.

Question 7. Evaluate \(\lim_{x \to \frac{\pi}{6}} \left( \frac{\cot^2 x - 3}{\csc x - 2} \right)\)
Answer: We have \(\lim_{x \to \frac{\pi}{6}} \left( \frac{\cot^2 x - 3}{\csc x - 2} \right)\)
\(= \lim_{x \to \frac{\pi}{6}} \left( \frac{(\csc^2 x - 1) - 3}{\csc x - 2} \right)\)
\(= \lim_{x \to \frac{\pi}{6}} \left( \frac{\csc^2 x - 4}{\csc x - 2} \right)\)
\(= \lim_{x \to \frac{\pi}{6}} \left( \frac{(\csc x + 2)(\csc x - 2)}{\csc x - 2} \right)\)
\(= \csc\left(\frac{\pi}{6}\right) + 2 = 2 + 2 = 4\)
In simple words: Convert \(\cot^2 x\) into \(\csc^2 x - 1\) and factor the resulting difference of squares to cancel the denominator.

Exam Tip: Convert all trigonometric functions to a single ratio (like cosecant) using Pythagorean identities to simplify rational trigonometric expressions.

Question 8. Evaluate \(\lim_{x \to 1} \left( \frac{x^7 - 2x^5 + 1}{x^3 - 3x^2 + 2} \right)\)
Answer: Since \(\lim_{x \to 1}\) gives \(\frac{0}{0}\), \((x - 1)\) is a factor of both numerator and denominator.
Dividing the numerator \(x^7 - 2x^5 + 1\) by \((x - 1)\) gives:
\(x^6 + x^5 - x^4 - x^3 - x^2 - x - 1\)
Dividing the denominator \(x^3 - 3x^2 + 2\) by \((x - 1)\) gives:
\(x^2 - 2x - 2\)
Therefore, the limit becomes:
\(\lim_{x \to 1} \frac{(x - 1)(x^6 + x^5 - x^4 - x^3 - x^2 - x - 1)}{(x - 1)(x^2 - 2x - 2)}\)
\(= \frac{1 + 1 - 1 - 1 - 1 - 1 - 1}{1 - 2 - 2}\)
\(= \frac{-5}{-3}\) Wait, recalculating numerator sum: \(1 + 1 - 1 - 1 - 1 - 1 - 1 = -3\).
\(= \frac{-3}{-3} = 1\)
In simple words: Factor out \((x - 1)\) from both numerator and denominator using polynomial division, then substitute \(x = 1\).

Exam Tip: For \(0/0\) indeterminate forms with polynomials, factor out \((x - a)\) using synthetic or long division.

Question 9. Evaluate \(\lim_{x \to 0} \left( \frac{|\sin x|}{x} \right)\)
Answer: First, find the Left-Hand Limit (L.H.L.):
\(\lim_{x \to 0^-} \left( \frac{|\sin x|}{x} \right)\)
Put \(x = 0 - h = -h\) and \(h \to 0\)
\(\lim_{h \to 0} \left( \frac{|\sin(-h)|}{-h} \right) = \lim_{h \to 0} \left( \frac{\sin h}{-h} \right) = -1\)
Now, find the Right-Hand Limit (R.H.L.):
\(\lim_{x \to 0^+} \left( \frac{|\sin x|}{x} \right)\)
Put \(x = 0 + h = h\) and \(h \to 0\)
\(\lim_{h \to 0} \left( \frac{|\sin h|}{h} \right) = \lim_{h \to 0} \left( \frac{\sin h}{h} \right) = 1\)
Since L.H.L. \(\neq\) R.H.L. (\(-1 \neq 1\)), \(\lim_{x \to 0} f(x)\) does not exist.
In simple words: Check the limit from the left and right separately; since they yield different values, the limit does not exist.

Exam Tip: Whenever modulus functions are involved, always evaluate left-hand and right-hand limits independently.