find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using…

find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it. lim x→1 (x^a - 1)/(x^b - 1), b ≠ 0

find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it. lim x→1 (x^a - 1)/(x^b - 1), b ≠ 0

Answer

Explanation:

Step1: Check form of limit

When (x\rightarrow1), we have (\lim_{x\rightarrow1}\frac{x^{a}-1}{x^{b}-1}=\frac{1^{a}-1}{1^{b}-1}=\frac{0}{0}), so L'Hopital's Rule can be applied.

Step2: Differentiate numerator and denominator

By the power - rule ((x^{n})^\prime=nx^{n - 1}), the derivative of (y = x^{a}-1) is (y^\prime=ax^{a - 1}), and the derivative of (y=x^{b}-1) is (y^\prime=bx^{b - 1}). So, (\lim_{x\rightarrow1}\frac{x^{a}-1}{x^{b}-1}=\lim_{x\rightarrow1}\frac{ax^{a - 1}}{bx^{b - 1}}).

Step3: Evaluate the new limit

Substitute (x = 1) into (\frac{ax^{a - 1}}{bx^{b - 1}}), we get (\frac{a\times1^{a - 1}}{b\times1^{b - 1}}=\frac{a}{b}).

Answer:

(\frac{a}{b})