limx→ 1[frac22x-2-2x+1sin²(x-1)]

Miscellaneous 7 II · Q131

\displaystyle\lim_{x\to 1}\left[\frac{2^{2x-2}-2^{x}+1}{\sin^2(x-1)}\right]

Maharashtra MsbshseTextbookSubjectiveImportance★★★★★est

24% · 33/137 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Let $u=x-1\to0$ . $2^{2x-2}=2^{2u}=(2^u)^2$ and $2^x=2^{1+u}=2\cdot2^u$ . So the numerator is $(2^u)^2-2\cdot2^u+1=(2^u-1)^2\sim(u\log2)^2=u^2(\log2)^2$ . Denominator $\sin^2 …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ130 — limx→ π/4[((sin x-cos x)²)/(√2-sin x-cos x)]→Next questionQ132 — limx→ 1[frac4x-1-2x+1(x-1)²]

Q. $\displaystyle\lim_{x\to 1}\left[\frac{2^{2x-2}-2^{x}+1}{\sin^2(x-1)}\right]$

Concept understanding — Limits of Exponential and Logarithmic Functions

Q.lim⁡x→1[22x−2−2x+1sin⁡2(x−1)]\displaystyle\lim_{x\to 1}\left[\frac{2^{2x-2}-2^{x}+1}{\sin^2(x-1)}\right]x→1lim​[sin2(x−1)22x−2−2x+1​]

🎓Unlock everything free for 14 days

Q. $\displaystyle\lim_{x\to 1}\left[\frac{2^{2x-2}-2^{x}+1}{\sin^2(x-1)}\right]$

Unlock everything free for 14 days