lim x tends to 0 [6^x-3^x-2^x+1/36^x-9^x-4^x+1]
Answers
Answer:
Applying L-Hospital's rule,
Again, applying L-Hospital's rule,
To solve the limit:
Lim x→0 [(6^x-3^x-2^x+1) / (36^x-9^x-4^x+1)]
We can factor the numerator and denominator by grouping the terms:
Lim x→0 [(6^x-3^x) - (2^x-1)] / [(36^x-9^x) - (4^x-1)]
Next, we can use the identities a^2 - b^2 = (a+b)(a-b) and a^3 - b^3 = (a-b)(a^2 + ab + b^2) to further simplify the expression:
Lim x→0 [3^x(2^x-1) - 1(2^x-1)] / [9^x(4^x-1) - 1(4^x-1)]
Lim x→0 [(2^x-1)(3^x-1)] / [(4^x-1)(9^x-1)]
Now, we can substitute x=0 to evaluate the limit:
(2^0-1)(3^0-1) / (4^0-1)(9^0-1) = (0)(0) / (0)(0) = indeterminate form
To determine the limit, we can use L'Hopital's rule by taking the derivative of the numerator and denominator with respect to x:
Lim x→0 [(ln6 * 6^x + ln3 * 3^x + ln2 * 2^x) - (ln2 * 2^x)] / [(ln36 * 36^x + ln9 * 9^x + ln4 * 4^x) - (ln4 * 4^x)]
Lim x→0 [(ln6 * 6^x + ln3 * 3^x) - (ln2 * 2^x)] / [(ln36 * 36^x + ln9 * 9^x) - (ln4 * 4^x)]
Lim x→0 [(ln6 * 6^x + ln3 * 3^x) - (ln2 * 2^x)] / [(ln6^2 * 6^x + ln3^2 * 3^x) - (ln2^2 * 2^x)]
Lim x→0 [(ln6 + ln3) - (ln2)] / [(ln6^2 + ln3^2) - (ln2^2)]
Now, we can substitute x=0 to evaluate the limit:
[(ln6 + ln3) - (ln2)] / [(ln6^2 + ln3^2) - (ln2^2)] = (ln(6*3) - ln2) / (ln(6^2) + ln(3^2) - ln(2^2))
= (ln18 - ln2) / (ln36 + ln9 - ln4)
= (ln9 - ln2) / (ln9 + ln2)
= (ln(9/2)) / (ln(9) + ln(2))
≈ 0.4619
Therefore, the limit of the expression is approximately 0.4619.
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