Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

Evaluate: $\underset{x→0}{\lim}\frac{27^x-9^x-3^x+1}{\sqrt{2}-\sqrt{1+\cos x}}$

Options:

$\sqrt{2}\log 3$

$8\sqrt{2}(\log 3)^2$

$8\sqrt{2}(\log 3)$

None of these

Correct Answer:

$8\sqrt{2}(\log 3)^2$

Explanation:

$\underset{x→0}{\lim}\frac{27^x-9^x-3^x+1}{\sqrt{2}-\sqrt{1+\cos x}}=\underset{x→0}{\lim}\frac{(3^x-1)(9^x-1)}{(1-\cos x)}(\sqrt{2}+\sqrt{1+\cos x})$ (Rationalise)

$\underset{x→0}{\lim}\frac{(3^x-1)(9^x-1)}{2\sin^2x/2}.\underset{x→0}{\lim}(\sqrt{2}+\sqrt{1+\cos x})=\underset{x→0}{\lim}\frac{\frac{3^x-1}{x}.\frac{9^x-1}{x}×\frac{2x^2}{4}.(\sqrt{2}+\sqrt{2})}{\sin^2x/2}$

$= 2ln\, 3 . ln\, 9 .2\sqrt{2}=8\sqrt{2}(ln\,3)^2$