Limits
Right-hand limit of complex expression at x=0
MJAT_TS8_P2
Grade 12
Question:
Let $f(x)=\begin{pmatrix}x^{\ln(2x-1)}\cdot\dfrac{(x)^{x\ln x}}{(x^{e^x-2}-1)\cdot x^e\cdot\sin x}\end{pmatrix}$. Then the right-hand limit of $f(x)$ at $x=0$ equals:
A) $\ln 2$
B) $e\ln 2$
C) $e^{\ln 2}$
D) does not exist
Step-by-Step Solution
Key Concept: Simplify the expression using $x^{\ln(2x-1)}=e^{\ln(2x-1)\ln x}$ and $x^e\cdot\sin x\sim x^{e+1}$ as $x\to 0^+$. Use standard limits $\lim_{x\to 0^+}\frac{x^{e^x-2}-1}{x}=\ln x\cdot\lim_{x\to 0^+}e^{x}=\ldots$
Limit $=e\ln 2$. Answer: **B**.
Correct Answer: B