Let $f(x) = \begin{cases} \frac{x^2 \sin\left(\frac{1}{x}\right) + 2x}{(1+x)^{\frac{1}{x}} - e}, & x \neq 0 \\ \lambda, & x = 0 \end{cases}$ If $f(x)$ is continuous at $x=0$, then the value of $\lambda$ is:
Step-by-Step Solution
Key Concept: For continuity at \(x=0\), compute
\[
\lambda=\lim_{x\to 0}
\frac{x^2\sin\left(\frac{1}{x}\right)+2x}
{(1+x)^{1/x}-e}.
\]
The term \(x^2\sin(1/x)\) is negligible compared with \(x\), so the numerator behaves like \(2x\).
For the denominator, use
\[
\frac{\ln(1+x)}{x}
=1-\frac{x}{2}+O(x^2),
\]
so
\[
(1+x)^{1/x}
=e^{1-x/2+O(x^2)}
=e\left(1-\frac{x}{2}+O(x^2)\right).
\]
Thus
\[
(1+x)^{1/x}-e\sim -\frac{ex}{2}.
\]
For \(f(x)\) to be continuous at \(x=0\), we need
\[
\lambda=\lim_{x\to 0}
\frac{x^2\sin\left(\frac{1}{x}\right)+2x}
{(1+x)^{1/x}-e}.
\]
First,
\[
x^2\sin\left(\frac{1}{x}\right)=O(x^2),
\]
so
\[
x^2\sin\left(\frac{1}{x}\right)+2x
=2x+O(x^2).
\]
Now consider the denominator. We have
\[
(1+x)^{1/x}
=e^{\ln(1+x)/x}.
\]
Using
\[
\ln(1+x)=x-\frac{x^2}{2}+O(x^3),
\]
we get
\[
\frac{\ln(1+x)}{x}
=1-\frac{x}{2}+O(x^2).
\]
Therefore
\[
(1+x)^{1/x}
=e^{1-x/2+O(x^2)}
=e\cdot e^{-x/2+O(x^2)}.
\]
Again expanding the exponential,
\[
e^{-x/2+O(x^2)}
=1-\frac{x}{2}+O(x^2).
\]
Hence
\[
(1+x)^{1/x}
=e\left(1-\frac{x}{2}+O(x^2)\right).
\]
So
\[
(1+x)^{1/x}-e
=-\frac{ex}{2}+O(x^2).
\]
Thus
\[
\lambda
=\lim_{x\to 0}
\frac{2x+O(x^2)}
{-\frac{ex}{2}+O(x^2)}
=-\frac{4}{e}.
\]
\[
\boxed{-\frac{4}{e}}
\]
Correct Answer: 4