Question

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

Explanation:

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

We know that \(\lim_{x
ightarrow0}(f(x)+ 1)^{\frac{1}{g(x)}}=\lim_{x
ightarrow0}e^{\frac{\ln(f(x)+1)}{g(x)}}\)
By the property of limits \(\lim_{x
ightarrow a}e^{h(x)}=e^{\lim_{x
ightarrow a}h(x)}\) (if \(\lim_{x
ightarrow a}h(x)\) exists), so we first find \(\lim_{x
ightarrow0}\frac{\ln(f(x)+1)}{g(x)}\)

Step2: Apply L - H rule

Since \(\lim_{x
ightarrow0}\ln(f(x)+1)=\ln(0 + 1)=0\) and \(\lim_{x
ightarrow0}g(x)=0\), 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 \(\lim_{x
ightarrow0}\frac{\ln(f(x)+1)}{g(x)}=\lim_{x
ightarrow0}\frac{\frac{f^\prime(x)}{f(x)+1}}{g^\prime(x)}\)

Step3: Substitute the limit values

Substitute \(\lim_{x
ightarrow0}f(x)=0\), \(\lim_{x
ightarrow0}f^\prime(x)=15\) and \(\lim_{x
ightarrow0}g^\prime(x)=5\) into \(\lim_{x
ightarrow0}\frac{\frac{f^\prime(x)}{f(x)+1}}{g^\prime(x)}\)
We get \(\frac{\frac{15}{0 + 1}}{5}=\frac{15}{5}=3\)

Answer:

\(e^{3}\)