limx→ 1[frac4x-1-2x+1(x-1)²]

Miscellaneous 7 II · Q132

\displaystyle\lim_{x\to 1}\left[\frac{4^{x-1}-2^{x}+1}{(x-1)^2}\right]

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Let $u=x-1\to0$ . $4^{x-1}=4^u=(2^u)^2$ and $2^x=2^{1+u}=2\cdot2^u$ . So the numerator is $(2^u)^2-2\cdot2^u+1=(2^u-1)^2\sim u^2(\log2)^2$ . Denominator …

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Q. $\displaystyle\lim_{x\to 1}\left[\frac{4^{x-1}-2^{x}+1}{(x-1)^2}\right]$

Concept understanding — Limits of Exponential and Logarithmic Functions

Q.lim⁡x→1[4x−1−2x+1(x−1)2]\displaystyle\lim_{x\to 1}\left[\frac{4^{x-1}-2^{x}+1}{(x-1)^2}\right]x→1lim​[(x−1)24x−1−2x+1​]

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Q. $\displaystyle\lim_{x\to 1}\left[\frac{4^{x-1}-2^{x}+1}{(x-1)^2}\right]$

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