question 10 (1 point)\nlet f and g be functions such that:\n$$\\lim_{x \\to 0} f(x) = 0, \\lim_{x \\to 0}…

question 10 (1 point)\nlet f and g be functions such that:\n$$\\lim_{x \\to 0} f(x) = 0, \\lim_{x \\to 0} f(x) = 15, \\lim_{x \\to 0} g(x) = 0, \\lim_{x \\to 0} g(x) = 5.$$\nthe limit\n$$\\lim_{x \\to 0} (f(x) + 1)^{1/g(x)}$$\nequals\n$$e^{75}$$\n$$\\infty$$\n$$e$$\n$$e^{3}$$\n$$1$$

question 10 (1 point)\nlet f and g be functions such that:\n$$\\lim_{x \\to 0} f(x) = 0, \\lim_{x \\to 0} f(x) = 15, \\lim_{x \\to 0} g(x) = 0, \\lim_{x \\to 0} g(x) = 5.$$\nthe limit\n$$\\lim_{x \\to 0} (f(x) + 1)^{1/g(x)}$$\nequals\n$$e^{75}$$\n$$\\infty$$\n$$e$$\n$$e^{3}$$\n$$1$$

Answer

Explanation:

Step1: Use the formula (a^b = e^{b\ln a})

We know that (\lim_{x\rightarrow0}(f(x)+ 1)^{\frac{1}{g(x)}}=\lim_{x\rightarrow0}e^{\frac{\ln(f(x)+1)}{g(x)}}) By the property of limits (\lim_{x\rightarrow a}e^{h(x)}=e^{\lim_{x\rightarrow a}h(x)}) (if (\lim_{x\rightarrow a}h(x)) exists), so we first find (\lim_{x\rightarrow0}\frac{\ln(f(x)+1)}{g(x)})

Step2: Apply L - H rule

Since (\lim_{x\rightarrow0}\ln(f(x)+1)=\ln(0 + 1)=0) and (\lim_{x\rightarrow0}g(x)=0), we can apply L - H rule. Differentiate the numerator and denominator: The derivative of (y = \ln(f(x)+1)) is (y^\prime=\frac{f^\prime(x)}{f(x)+1}), and the derivative of (y = g(x)) is (y^\prime=g^\prime(x)) So (\lim_{x\rightarrow0}\frac{\ln(f(x)+1)}{g(x)}=\lim_{x\rightarrow0}\frac{\frac{f^\prime(x)}{f(x)+1}}{g^\prime(x)})

Step3: Substitute the limit values

Substitute (\lim_{x\rightarrow0}f(x)=0), (\lim_{x\rightarrow0}f^\prime(x)=15) and (\lim_{x\rightarrow0}g^\prime(x)=5) into (\lim_{x\rightarrow0}\frac{\frac{f^\prime(x)}{f(x)+1}}{g^\prime(x)}) We get (\frac{\frac{15}{0 + 1}}{5}=\frac{15}{5}=3)

Answer:

(e^{3})