question 2: limit evaluation\nevaluate the following limit. if the limit does not exist, enter dne. if the…

question 2: limit evaluation\nevaluate the following limit. if the limit does not exist, enter dne. if the limit is infinite, specify if it is positive or negative infinity.\n lim_{x\rightarrow5^{+}}\frac{ln(x)-ln(3)}{ln(x)-ln(5)} \nplease simplify your answer.\n lim_{x\rightarrow5^{+}}\frac{ln(x)-ln(3)}{ln(x)-ln(5)}=\text{symbolic expression}

question 2: limit evaluation\nevaluate the following limit. if the limit does not exist, enter dne. if the limit is infinite, specify if it is positive or negative infinity.\n lim_{x\rightarrow5^{+}}\frac{ln(x)-ln(3)}{ln(x)-ln(5)} \nplease simplify your answer.\n lim_{x\rightarrow5^{+}}\frac{ln(x)-ln(3)}{ln(x)-ln(5)}=\text{symbolic expression}

Answer

Explanation:

Step1: Use logarithmic property

Recall that $\ln a-\ln b=\ln\frac{a}{b}$. So, $\lim_{x\rightarrow5^{+}}\frac{\ln(x)-\ln(3)}{\ln(x)-\ln(5)}=\lim_{x\rightarrow5^{+}}\frac{\ln\frac{x}{3}}{\ln\frac{x}{5}}$.

Step2: Substitute $x = 5^{+}$

As $x\rightarrow5^{+}$, we have $\frac{\ln\frac{x}{3}}{\ln\frac{x}{5}}$. When $x\rightarrow5^{+}$, $\ln\frac{x}{3}\rightarrow\ln\frac{5}{3}$ and $\ln\frac{x}{5}\rightarrow0^{+}$. Since $\ln\frac{5}{3}>0$ and $\ln\frac{x}{5}\rightarrow0^{+}$ as $x\rightarrow5^{+}$, the value of the limit is $+\infty$.

Answer:

$+\infty$