evaluate the following limit. use lhôpitals rule when it is convenient and applicable.\n lim_{x\rightarrowinf…

Explanation:

Step1: Let (y=(1 + \frac{a}{x})^x)

Take the natural - logarithm of both sides: (\ln y=x\ln(1 + \frac{a}{x}))

Step2: Rewrite the limit

We want to find (\lim_{x\rightarrow\infty}y), which is equivalent to finding (\lim_{x\rightarrow\infty}\ln y) first. So, (\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}x\ln(1 + \frac{a}{x})=\lim_{x\rightarrow\infty}\frac{\ln(1+\frac{a}{x})}{\frac{1}{x}}) As (x\rightarrow\infty), we have the (\frac{0}{0}) indeterminate form.

Step3: Apply L'Hopital's Rule

Differentiate the numerator and the denominator. The derivative of (\ln(1+\frac{a}{x})) with respect to (x) is (\frac{1}{1 + \frac{a}{x}}\cdot(-\frac{a}{x^{2}})), and the derivative of (\frac{1}{x}) with respect to (x) is (-\frac{1}{x^{2}}). So, (\lim_{x\rightarrow\infty}\frac{\ln(1+\frac{a}{x})}{\frac{1}{x}}=\lim_{x\rightarrow\infty}\frac{\frac{1}{1+\frac{a}{x}}\cdot(-\frac{a}{x^{2}})}{-\frac{1}{x^{2}}}=\lim_{x\rightarrow\infty}\frac{a}{1+\frac{a}{x}})

Step4: Evaluate the limit

As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{a}{1+\frac{a}{x}} = a) Since (\lim_{x\rightarrow\infty}\ln y=a), then (\lim_{x\rightarrow\infty}y = e^{a})

Answer:

(e^{a})