Determine the value of the limit $\lim _ { x \rightarrow 0 ^ { + } } \frac { \ln x } { \ln \left( x ^ { 2 } + 2 x \right) }$ ?
A) $- \infty$
B)-1
C)1
D)0
E) $\infty$

Question

Quizplus · Accepted Answer

As $ x \to 0^+ $, both $\ln x$ and $\ln(x^2 + 2x)$ approach $-\infty$. Using L'Hôpital's Rule, differentiate the numerator and denominator: $\frac{1/x}{(2x+2)/(x^2+2x)}$. Simplifying gives $\frac{x^2+2x}{2x^2+2x}$, which approaches 1 as $ x \to 0^+ $. Thus, the limit is 1.

Determine the Value of the Limit lim⁡x→0+ln⁡xln⁡(x2+2x)\lim _ { x \rightarrow 0 ^ { + } } \frac { \ln x } { \ln \left( x ^ { 2 } + 2 x \right) }limx→0+​ln(x2+2x)lnx​

Determine the Value of the Limit $\lim _ { x \rightarrow 0 ^ { + } } \frac { \ln x } { \ln \left( x ^ { 2 } + 2 x \right) }$