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How to calculate this LIMIT?
Limit x-> 0(+) (8^x - 4^x - 2^x +1) / x^2.
I tried using l hospital too but in vain, please help me. Thank you. BA for grabs.
3 Answers
- ?Lv 75 years agoFavorite Answer
d/dx (a^x) = (ln a) a^x
lim[x→0⁺] (8^x − 4^x − 2^x + 1) / x^2
= lim[x→0⁺] (2^(3x) − 2^(2x) − 2^x + 1) / x^2
= 0/0
Use L'Hopital's Rule
= lim[x→0⁺] (ln 2) (3 * 2^(3x) − 2 * 2^(2x) − 2^x) / (2x)
= (ln 2) (3−2−1) / 0
= 0/0
Use L'Hopital's Rule again
= lim[x→0⁺] (ln 2)² (9 * 2^(3x) − 4 * 2^(2x) − 2^x) / 2
= (ln 2)² (9−4−1)/2
= 2 (ln 2)²
≈ 0.96090602783640284933420505265333
Check:
- MorningfoxLv 75 years ago
If you plot it from (for example) from -0.1 to +0.1, you'll see that there is no special behavior at x = 0. You can approach the limit from either side, and use L'Hospital's rule as many times as needed.
