Limits, Continuity & Differentiability
Non-existence of Limits and One-sided Limits
Grade 12
Question:
<p><strong>Ex. 38:</strong> Statement I: If $\lim_{x \to 0} \frac{f(x)}{\sin x}$ does not exist, then $\lim_{x \to 0} f(x)$ does not exist.</p><p>Statement II: $\lim_{x \to 0} \frac{e^{1/x} - 1}{e^{1/x} + 1}$ does not exist.</p>
<p>(a) Statement I is true, Statement II is true; Statement II is correct explanation for Statement I</p>
<p>(b) Statement I is true, Statement II is true; Statement II is not the correct explanation for Statement I</p>
<p>(c) Statement I is true, Statement II is false</p>
<p>(d) Statement I is false, Statement II is true</p>
Step-by-Step Solution
Key Concept: Non-existence of $\lim \frac{f(x)}{\sin x}$ does not necessarily imply non-existence of $\lim f(x)$; check limits by examining left and right limits separately.
<p><strong>Solution:</strong> Consider Statement I. If $\lim_{x \to 0} f(x) = L$ exists (and is finite), then $\lim_{x \to 0} \frac{f(x)}{\sin x} = \lim_{x \to 0} f(x) \cdot \frac{1}{\sin x}$. If $L \neq 0$, this limit does not exist. However, if $L = 0$, the limit might exist or not depending on the rate of decay of $f(x)$. Counterexample: $f(x) = x^2$. Then $\lim_{x \to 0} \frac{x^2}{\sin x} = \lim_{x \to 0} x \cdot \frac{x}{\sin x} = 0$ (exists), but we can construct $f(x)$ such that the first limit doesn't exist while the second does. Statement I is false. For Statement II, as $x \to 0^+$, $e^{1/x} \to \infty$, giving $\frac{e^{1/x}-1}{e^{1/x}+1} \to 1$. As $x \to 0^-$, $e^{1/x} \to 0$, giving $\frac{e^{1/x}-1}{e^{1/x}+1} \to -1$. Left and right limits differ, so the limit does not exist. Statement II is true.</p><p>∴ Answer is (d): Statement I is false, Statement II is true.</p>
Correct Answer: D