Question

evaluate the following limit. use lhôpitals rule when it is convenient and applicable.
lim_{x
ightarrowinfty}left(1 + \frac{a}{x}
ight)^{x}, \text{ for some constant } a
lim_{x
ightarrowinfty}left(1 + \frac{a}{x}
ight)^{x}=square \text{ (type an exact answer.)}

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 \(\lim_{x
ightarrow\infty}y\), which is equivalent to finding \(\lim_{x
ightarrow\infty}\ln y\) first. So, \(\lim_{x
ightarrow\infty}\ln y=\lim_{x
ightarrow\infty}x\ln(1 + \frac{a}{x})=\lim_{x
ightarrow\infty}\frac{\ln(1+\frac{a}{x})}{\frac{1}{x}}\)
As \(x
ightarrow\infty\), 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, \(\lim_{x
ightarrow\infty}\frac{\ln(1+\frac{a}{x})}{\frac{1}{x}}=\lim_{x
ightarrow\infty}\frac{\frac{1}{1+\frac{a}{x}}\cdot(-\frac{a}{x^{2}})}{-\frac{1}{x^{2}}}=\lim_{x
ightarrow\infty}\frac{a}{1+\frac{a}{x}}\)

Step4: Evaluate the limit

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

Answer:

\(e^{a}\)