Ex 5.9 – Continuity and Differentiability | ML Aggarwal Class 12 Solutions

Here is the complete ML Aggarwal Class 12 Solutions of Exercise – 5.9 for Chapter 5 – Continuity and Differentiability. Each question is solved step by step for better understanding.

Question: 1. (i) Evaluate \(\lim_{x \to \infty} \frac{1}{(x-3)^2}\).
Solution: We note that as \( x \to \infty \), the term \(\,(x-3)^2\) grows without bound. Hence, \[ \lim_{x \to \infty} (x-3)^2 = \infty. \] Consequently, \[ \lim_{x \to \infty} \frac{1}{(x-3)^2} = \frac{1}{\infty} = 0. \]
Final Answer: \(\lim_{x \to \infty} \frac{1}{(x-3)^2} = 0\).

Question: 1. (ii) Evaluate \(\lim_{x \to \infty} \frac{1}{(1-x)^2}\).
Solution: As \(x \to \infty\), the dominant term in \((1-x)^2\) is \(x^2\) (since \((-x)^2 = x^2\)). Thus, \[ \lim_{x \to \infty} (1-x)^2 = \infty. \] Consequently, \[ \lim_{x \to \infty} \frac{1}{(1-x)^2} = \frac{1}{\infty} = 0. \]
Final Answer: \(\lim_{x \to \infty} \frac{1}{(1-x)^2} = 0\).

Question: 2. (i) Evaluate \(\lim_{x \to \infty} \cos x\).
Solution: The cosine function, \(\cos x\), oscillates between \(-1\) and \(1\) for all \(x\). As \(x \to \infty\), \(\cos x\) does not settle at a single value but continues to oscillate. Therefore, \[ \lim_{x \to \infty} \cos x \quad \text{does not exist}. \]
Final Answer: \(\lim_{x \to \infty} \cos x\) does not exist.

Question: 2. (ii) Evaluate \(\lim_{x \to 0} \cos \frac{1}{x}\).
Solution: As \(x \to 0\), the expression \(\frac{1}{x}\) becomes unbounded, leading to rapid oscillations in \(\cos \frac{1}{x}\). Since the cosine function oscillates between \(-1\) and \(1\) without approaching a unique value, \[ \lim_{x \to 0} \cos \frac{1}{x} \quad \text{does not exist}. \]
Final Answer: \(\lim_{x \to 0} \cos \frac{1}{x}\) does not exist.

Question: 3. (i) Evaluate \(\lim_{x \to 0} \frac{e^{-x}-1}{x}\).
Solution: Using the Taylor series expansion of \(e^{-x}\) about \(x=0\): \[ e^{-x} = 1 – x + \frac{x^2}{2} – \frac{x^3}{6} + \cdots, \] we have: \[ e^{-x}-1 = -x + \frac{x^2}{2} – \frac{x^3}{6} + \cdots. \] Dividing by \(x\): \[ \frac{e^{-x}-1}{x} = -1 + \frac{x}{2} – \frac{x^2}{6} + \cdots. \] Taking the limit as \(x \to 0\), all higher order terms vanish, yielding: \[ \lim_{x \to 0} \frac{e^{-x}-1}{x} = -1. \]
Final Answer: \(\lim_{x \to 0} \frac{e^{-x}-1}{x} = -1\).

Question: 3. (ii) Evaluate \(\lim_{x \to 0} \frac{e^x – e^{-x}}{x}\).
Solution: Expanding \(e^x\) and \(e^{-x}\) using their Taylor series about \(x=0\): \[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots, \] \[ e^{-x} = 1 – x + \frac{x^2}{2} – \frac{x^3}{6} + \cdots. \] Subtracting the series: \[ e^x – e^{-x} = (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots) – (1 – x + \frac{x^2}{2} – \frac{x^3}{6} + \cdots) = 2x + \frac{2x^3}{6} + \cdots. \] Dividing by \(x\): \[ \frac{e^x – e^{-x}}{x} = 2 + \frac{2x^2}{6} + \cdots. \] Taking the limit as \(x \to 0\), the higher order terms vanish, giving: \[ \lim_{x \to 0} \frac{e^x – e^{-x}}{x} = 2. \]
Final Answer: \(\lim_{x \to 0} \frac{e^x – e^{-x}}{x} = 2\).

Question: 4. (i) Evaluate \(\lim_{x \to 0} \frac{7^x – 1}{\tan x}\).
Solution: Express \(7^x\) in exponential form: \[ 7^x = e^{x\,\text{log}\,7}. \] Thus, \[ 7^x – 1 = e^{x\,\text{log}\,7} – 1. \] Using the Taylor expansion for \(e^u\) at \(u = 0\) with \(u = x\,\text{log}\,7\), we have: \[ e^{x\,\text{log}\,7} = 1 + x\,\text{log}\,7 + O(x^2). \] Hence, \[ 7^x – 1 \approx x\,\text{log}\,7. \] Since \(\tan x \approx x\) for small \(x\), we have: \[ \lim_{x \to 0} \frac{7^x – 1}{\tan x} \approx \lim_{x \to 0} \frac{x\,\text{log}\,7}{x} = \text{log}\,7. \]
Final Answer: \(\lim_{x \to 0} \frac{7^x – 1}{\tan x} = \text{log}\,7\).

Question: 4. (ii) Evaluate \(\lim_{x \to 0} \frac{3^x – 1}{\sqrt{2+x}-\sqrt{2}}\).
Solution: Express \(3^x\) as: \[ 3^x = e^{x\,\text{log}\,3}. \] For small \(x\), using the Taylor expansion, \[ 3^x – 1 \approx x\,\text{log}\,3. \] Next, expand \(\sqrt{2+x}\) using the binomial expansion: \[ \sqrt{2+x} = \sqrt{2}\sqrt{1+\frac{x}{2}} \approx \sqrt{2}\left(1+\frac{x}{4}\right). \] Thus, \[ \sqrt{2+x} – \sqrt{2} \approx \sqrt{2}\cdot\frac{x}{4} = \frac{x\sqrt{2}}{4}. \] The limit becomes: \[ \lim_{x \to 0} \frac{3^x – 1}{\sqrt{2+x}-\sqrt{2}} \approx \lim_{x \to 0} \frac{x\,\text{log}\,3}{\frac{x\sqrt{2}}{4}} = \frac{4\,\text{log}\,3}{\sqrt{2}} = 2\sqrt{2}\,\text{log}\,3. \]
Final Answer: \(\lim_{x \to 0} \frac{3^x – 1}{\sqrt{2+x}-\sqrt{2}} = 2\sqrt{2}\,\text{log}\,3\).

Question: 5. (i) Evaluate \(\lim_{x \to 0} \frac{e^x – x – 1}{x}\).
Solution: Using the Taylor series expansion for \(e^x\): \[ e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots, \] we get: \[ e^x – x – 1 = \frac{x^2}{2} + \frac{x^3}{6} + \cdots. \] Dividing by \(x\): \[ \frac{e^x – x – 1}{x} = \frac{x}{2} + \frac{x^2}{6} + \cdots. \] Taking the limit as \(x \to 0\), all terms vanish, thus: \[ \lim_{x \to 0} \frac{e^x – x – 1}{x} = 0. \]
Final Answer: \(\lim_{x \to 0} \frac{e^x – x – 1}{x} = 0\).

Question: 5. (ii) Evaluate \(\lim_{x \to 1} \frac{x-1}{\log x}\).
Solution: We observe that as \(x \to 1\), both the numerator and the denominator tend to 0. Applying L’Hôpital’s Rule, differentiate the numerator and denominator: \[ \frac{d}{dx}(x-1) = 1, \quad \frac{d}{dx}(\log x) = \frac{1}{x}. \] Hence, the limit becomes: \[ \lim_{x \to 1} \frac{1}{\frac{1}{x}} = \lim_{x \to 1} x = 1. \]
Final Answer: \(\lim_{x \to 1} \frac{x-1}{\log x} = 1\).

Question 6: Examine the function \[ f(x) = \begin{cases} e^{1/x}, & x \ne 0 \\ 1, & x = 0 \end{cases} \] for continuity at \( x = 0 \).
Solution: To check the continuity of \( f(x) \) at \( x = 0 \), we examine the left-hand and right-hand limits. For \( x \to 0^+ \): As \( x \to 0^+ \), \( \frac{1}{x} \to +\infty \), so \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^{1/x} = \infty. \] For \( x \to 0^- \): As \( x \to 0^- \), \( \frac{1}{x} \to -\infty \), so \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} e^{1/x} = 0. \] Since the left-hand and right-hand limits are not equal, the limit \( \lim_{x \to 0} f(x) \) does not exist. Also, \( f(0) = 1 \), so we cannot define continuity at \( x = 0 \). Hence, the function is not continuous at \( x = 0 \).
Final Answer: The function is discontinuous at \( x = 0 \).

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