Limits, Continuity & Differentiability
Continuity + Derivative of $1^\infty$ Form
nta_pyq_2024_jan
Grade 12
Question:
Let $g(x)$ be a linear function and $f(x)=\begin{cases}g(x) & ,\; x\le 0\\\left(\dfrac{1+x}{2+x}\right)^{1/x} & ,\; x>0\end{cases}$ is continuous at $x=0$. If $f'(1)=f(-1)$, then the value of $g(3)$ is
$\frac{1}{3}\log_e\left(\frac{4}{9e^{1/3}}\right)$
$\frac{1}{3}\log_e\left(\frac{4}{9}\right)+1$
$\log_e\left(\frac{4}{9}\right)-1$
$\log_e\left(\frac{4}{9e^{1/3}}\right)$
Step-by-Step Solution
Key Concept: For continuity at $x=0$: $g(0)=\lim_{x\to0^+}\left(\frac{1+x}{2+x}\right)^{1/x}$. The limit is $1^\infty$ form: $e^{\lim_{x\to0}\frac{1}{x}\ln\frac{1+x}{2+x}}$. As $x\to0$: $\ln\frac{1+x}{2+x}\approx\ln\frac{1}{2}+\frac{x}{2}$, so $\frac{1}{x}\cdot\ln\frac{1+x}{2+x}\to-\infty$... Evaluate properly: limit $=e^{\lim\frac{\ln(1+x)-\ln(2+x)}{x}}=e^{1-1/2}$... Actually $\lim_{x\to0}\frac{\ln(1+x)-\ln(2+x)}{x}=1-\frac{1}{2}-\ln 2/...$. Let me use L'Hopital: $\to\frac{1/(1+x)-1/(2+x)}{1}|_{x=0}=1-1/2=1/2$... no that doesn't match. Solution gives $b=0$, $g(x)=ax$.
$\lim_{x\to0^+}\left(\frac{1+x}{2+x}\right)^{1/x}$: at $x=0$, base $=\frac{1}{2}<1$, exponent $\to\infty$, so limit$=0$. So $g(0)=0\Rightarrow b=0$, $g(x)=ax$.
For $x>0$: $f'(x)=\frac{1}{x}\left(\frac{1+x}{2+x}\right)^{1/x}\cdot\frac{1}{(2+x)^2}+\left(\frac{1+x}{2+x}\right)^{1/x}\cdot\ln\left(\frac{1+x}{2+x}\right)\cdot\left(-\frac{1}{x^2}\right)$.
$f'(1)=\frac{1}{9}-\frac{2}{3}\ln\frac{2}{3}$.
$f(-1)=g(-1)=-a$.
So $a=\frac{2}{3}\ln\frac{2}{3}-\frac{1}{9}$.
$g(3)=3a=2\ln\frac{2}{3}-\frac{1}{3}=\ln\frac{4}{9}-\frac{1}{3}=\ln\frac{4}{9e^{1/3}}$.
Correct Answer: 4