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 \( x = 0 \).