NIMCET 2027 Limit Previous Year Questions (PYQs) – Page 1 of 3
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Solution
Solution
Quick DL Method Solution
Given:
\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \]
LHS using derivative:
\[ \lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \left.\frac{d}{dx}(x^4)\right|_{x=1} = 4x^3|_{x=1} = 4 \]
RHS using DL logic:
\[ \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \approx \frac{3k^2(x - k)}{2k(x - k)} = \frac{3k}{2} \]
Equating both sides:
\[ \frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3} \]
\[ \boxed{k = \frac{8}{3}} \]
Solution
Solution
Solution
Solution
Rationalize numerator and denominator:
\[ \frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}} =\frac{\dfrac{3(x-3)}{\sqrt{3x}+3}}{\dfrac{2(x-3)}{\sqrt{2x-4}+\sqrt{2}}} =\frac{3}{\sqrt{3x}+3}\cdot\frac{\sqrt{2x-4}+\sqrt{2}}{2}. \]
Now let \(x\to 3\): \(\sqrt{3x}\to 3\) and \(\sqrt{2x-4}\to \sqrt{2}\). Hence
\[ \lim_{x\to 3}=\frac{3}{3+3}\cdot\frac{\sqrt{2}+\sqrt{2}}{2} =\frac{1}{2}\cdot\sqrt{2}=\boxed{\frac{1}{2\sqrt{2}}}. \]
Solution
Solution
Solution
NIMCET 2027
Online Test Series, Information About Examination,
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and More.
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NIMCET 2027
Online Test Series, Information About Examination,
Syllabus, Notification
and More.
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