Question

lhopital: problem 1
(1 point)
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evaluate the limit using lhospitals rule if necessary
\\( \lim _ { x \
ightarrow \infty } \left( 1 + \frac { 10 } { x } \
ight) ^ { x } \\)
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Explanation:

Step1: Recognize the limit form

We know that the limit \(\lim_{x
ightarrow\infty}(1 + \frac{a}{x})^{bx}\) is of the form \(1^{\infty}\). We can use the formula \(y=(1+\frac{10}{x})^{x}\), and then take the natural logarithm. Let \(L=\lim_{x
ightarrow\infty}(1 + \frac{10}{x})^{x}\). Take \(\ln L=\lim_{x
ightarrow\infty}x\ln(1+\frac{10}{x})\).

Step2: Rewrite the limit

Rewrite \(\lim_{x
ightarrow\infty}x\ln(1+\frac{10}{x})\) as \(\lim_{x
ightarrow\infty}\frac{\ln(1 + \frac{10}{x})}{\frac{1}{x}}\). Now, as \(x
ightarrow\infty\), we have the \(\frac{0}{0}\) form.

Step3: Apply L'Hospital's Rule

Differentiate the numerator and denominator. The derivative of \(y = \ln(1+\frac{10}{x})\) is \(y^\prime=\frac{1}{1+\frac{10}{x}}\times(-\frac{10}{x^{2}})\), and the derivative of \(y=\frac{1}{x}\) is \(y^\prime=-\frac{1}{x^{2}}\). Then \(\lim_{x
ightarrow\infty}\frac{\frac{1}{1+\frac{10}{x}}\times(-\frac{10}{x^{2}})}{-\frac{1}{x^{2}}}=\lim_{x
ightarrow\infty}\frac{10}{1+\frac{10}{x}}\).

Step4: Evaluate the limit

As \(x
ightarrow\infty\), \(\lim_{x
ightarrow\infty}\frac{10}{1+\frac{10}{x}} = 10\). Since \(\ln L = 10\), then \(L = e^{10}\).

Answer:

\(e^{10}\)