Question

(2) \( \lim _{x \rightarrow 2} \frac{1-\cos ^{2}(x-2)}{(x-2)}=\lim _{x \rightarrow 2} \) (3) \( \lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{x}=7 \) (4) \( \lim _{x \rightarrow 1} \frac{\ln \left(8 x^{3}\right)-\ln 8}{\sin (\ln x)}\left(\frac{0}{0}\right) \)

          (2) \( \lim _{x \rightarrow 2} \frac{1-\cos ^{2}(x-2)}{(x-2)}=\lim _{x \rightarrow 2} \)
(3) \( \lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{x}=7 \)
(4) \( \lim _{x \rightarrow 1} \frac{\ln \left(8 x^{3}\right)-\ln 8}{\sin (\ln x)}\left(\frac{0}{0}\right) \)

(2) limx → 2(1-cos ^2(x-2))/((x-2))=limx → 2
(3) limx → 0(1-√(cos x))/(x)=7
(4) limx → 1(ln(8 x^3)-ln 8)/(sin (ln x))((0)/(0))

Added by Anna T.

Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Sanjiva Dayal IIT Kanpur 1st Edition

Chapter 2

Instant Answer

Solved by Expert Madhur L

08/10/2023

Step 1

### Problem 2: \[ \lim_{x \rightarrow 2} \frac{1-\cos^2(x-2)}{x-2} \] Show more…

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(2) \( \lim _{x \rightarrow 2} \frac{1-\cos ^{2}(x-2)}{(x-2)}=\lim _{x \rightarrow 2} \) (3) \( \lim _{x \rightarrow 0} \frac{1-\sqrt{\cos x}}{x}=7 \) (4) \( \lim _{x \rightarrow 1} \frac{\ln \left(8 x^{3}\right)-\ln 8}{\sin (\ln x)}\left(\frac{0}{0}\right) \)