lhopital: problem 1\n(1 point)\nresults for this submission\nthe answer is not correct.\nevaluate the limit…
lhopital: problem 1\n(1 point)\nresults for this submission\nthe answer is not correct.\nevaluate the limit using lhospitals rule if necessary\n\\( \\lim _ { x \\rightarrow \\infty } \\left( 1 + \\frac { 10 } { x } \\right) ^ { x } \\)\npreview my answers submit answers\nyour score was recorded.\nyour score was successfully sent to canvas.\nyou have attempted this problem 1 time.\nyou received a score of 0% for this attempt.\nyour overall recorded score is 0%.\nyou have unlimited attempts remaining.\nemail instructor\npage generated october 29, 2025 at 11:58:27 am cdt\nwebwork © 1996 - 2024 | theme: math4 | ww_version: 2.19 | pg_version\nthe webwork project
Answer
Explanation:
Step1: Recognize the limit form
We know that the limit (\lim_{x\rightarrow\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\rightarrow\infty}(1 + \frac{10}{x})^{x}). Take (\ln L=\lim_{x\rightarrow\infty}x\ln(1+\frac{10}{x})).
Step2: Rewrite the limit
Rewrite (\lim_{x\rightarrow\infty}x\ln(1+\frac{10}{x})) as (\lim_{x\rightarrow\infty}\frac{\ln(1 + \frac{10}{x})}{\frac{1}{x}}). Now, as (x\rightarrow\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\rightarrow\infty}\frac{\frac{1}{1+\frac{10}{x}}\times(-\frac{10}{x^{2}})}{-\frac{1}{x^{2}}}=\lim_{x\rightarrow\infty}\frac{10}{1+\frac{10}{x}}).
Step4: Evaluate the limit
As (x\rightarrow\infty), (\lim_{x\rightarrow\infty}\frac{10}{1+\frac{10}{x}} = 10). Since (\ln L = 10), then (L = e^{10}).
Answer:
(e^{10})