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. \n\nlim_{x\\to1^+}\\ln(x^5 - 1)-\\ln(x^3 - 1)

find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using it. \n\nlim_{x\\to1^+}\\ln(x^5 - 1)-\\ln(x^3 - 1)

Answer

Explanation:

Step1: Use logarithmic property

Use the property (\ln a-\ln b = \ln\frac{a}{b}). So, (\lim_{x\rightarrow1^{+}}[\ln(x^{5}-1)-\ln(x^{3}-1)]=\lim_{x\rightarrow1^{+}}\ln\frac{x^{5}-1}{x^{3}-1})

Step2: Factor the numerator and denominator

Factor (x^{n}-1=(x - 1)(x^{n - 1}+x^{n - 2}+\cdots+1)). Then (x^{5}-1=(x - 1)(x^{4}+x^{3}+x^{2}+x + 1)) and (x^{3}-1=(x - 1)(x^{2}+x + 1)) So, (\frac{x^{5}-1}{x^{3}-1}=\frac{(x - 1)(x^{4}+x^{3}+x^{2}+x + 1)}{(x - 1)(x^{2}+x + 1)}=\frac{x^{4}+x^{3}+x^{2}+x + 1}{x^{2}+x + 1}) for (x\neq1)

Step3: Find the limit

Now, (\lim_{x\rightarrow1^{+}}\ln\frac{x^{5}-1}{x^{3}-1}=\ln\left(\lim_{x\rightarrow1^{+}}\frac{x^{4}+x^{3}+x^{2}+x + 1}{x^{2}+x + 1}\right)) Substitute (x = 1) into (\frac{x^{4}+x^{3}+x^{2}+x + 1}{x^{2}+x + 1}), we get (\frac{1+1+1+1+1}{1+1+1}=\frac{5}{3})

Answer:

(\ln\frac{5}{3})