Question

use lhôpitals rule to find the limit.

$$\lim_{x \to 0^{+}} \frac{\ln (x^{2}+4x)}{\ln x}$$

$$\lim_{x \to 0^{+}} \frac{\ln (x^{2}+4x)}{\ln x}=\square$$ (type an exact answer.)

Explanation:

Step1: Check the form of the limit

As \(x\to0^{+}\), \(\ln(x^{2}+4x)\to-\infty\) and \(\ln x\to-\infty\). So, it is in the \(\frac{-\infty}{-\infty}\) form, and we can apply L'Hôpital's Rule.
By L'Hôpital's Rule, if \(\lim_{x\to a}\frac{f(x)}{g(x)}\) is in \(\frac{0}{0}\) or \(\frac{\pm\infty}{\pm\infty}\) form, then \(\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}\)
Let \(f(x)=\ln(x^{2}+4x)\) and \(g(x)=\ln x\)
First, find the derivatives:
\(f^{\prime}(x)=\frac{2x + 4}{x^{2}+4x}\) (using the chain - rule \((\ln u)^{\prime}=\frac{u^{\prime}}{u}\), where \(u = x^{2}+4x\) and \(u^{\prime}=2x + 4\))
\(g^{\prime}(x)=\frac{1}{x}\) (since \((\ln x)^{\prime}=\frac{1}{x}\))

Step2: Apply L'Hôpital's Rule

\(\lim_{x\to0^{+}}\frac{\ln(x^{2}+4x)}{\ln x}=\lim_{x\to0^{+}}\frac{\frac{2x + 4}{x^{2}+4x}}{\frac{1}{x}}\)
Simplify the complex fraction: \(\lim_{x\to0^{+}}\frac{(2x + 4)x}{x^{2}+4x}=\lim_{x\to0^{+}}\frac{2x^{2}+4x}{x^{2}+4x}\)

Step3: Simplify the function

Factor the numerator and denominator:
\(\lim_{x\to0^{+}}\frac{x(2x + 4)}{x(x + 4)}\)
Cancel out the non - zero factor \(x\) (since \(x\to0^{+}\), \(x
eq0\))
\(\lim_{x\to0^{+}}\frac{2x + 4}{x + 4}\)

Step4: Evaluate the limit

Substitute \(x = 0\) into \(\frac{2x+4}{x + 4}\)
\(\frac{2\times0+4}{0 + 4}=1\)

Answer:

\(1\)