Question:

Step 1: State the condition for continuity. For $f(x)$ to be continuous at $x = 0$, the function value at $x=0$ must be equal to the limit of the function as $x \to 0$. $$ \lambda = \lim_{x\to 0}\frac{x^2\sin\left(\frac{1}{x}\right)+2x}{(1+x)^{1/x}-e} $$ Step 2: Analyze the numerator as $x \to 0$. As $x \to 0$, the term $x^2\sin\left(\frac{1}{x}\right)$ approaches $0$ because $x^2$ is infinitesimal and $\sin\left(\frac{1}{x}\right)$ is bounded between $-1$ and $1$. Therefore, the numerator can be approximated by its leading term: $$ \text{Numerator} = 2x + x^2\sin\left(\frac{1}{x}\right) $$ Step 3: Expand $(1+x)^{1/x}$ using logarithms and Taylor series. We can rewrite $(1+x)^{1/x}$ using the exponential function: $$ (1+x)^{1/x} = e^{\frac{1}{x}\ln(1+x)} $$ Using the Taylor expansion for $\ln(1+x)$ around $x=0$: $$ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - O(x^4) $$ Dividing by $x$: $$ \frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3} - O(x^3) $$ Step 4: Simplify the exponent. Substitute the expansion into the exponential expression: $$ (1+x)^{1/x} = e^{1 - \frac{x}{2} + \frac{x^2}{3} - O(x^3)} = e \cdot e^{-\frac{x}{2} + \frac{x^2}{3} - O(x^3)} $$ Step 5: Analyze the denominator as $x \to 0$. Using the Taylor expansion $e^u = 1 + u + O(u^2)$ for small $u$, let $u = -\frac{x}{2} + \frac{x^2}{3} - O(x^3)$: $$ e^{-\frac{x}{2} + \frac{x^2}{3} - O(x^3)} = 1 + \left(-\frac{x}{2} + \frac{x^2}{3}\right) + O(x^2) = 1 - \frac{x}{2} + O(x^2) $$ Now, substitute this back into the denominator expression: $$ (1+x)^{1/x} - e = e\left(e^{-\frac{x}{2}+O(x^2)}-1\right) = e\left(1 - \frac{x}{2} + O(x^2) - 1\right) = e\left(-\frac{x}{2} + O(x^2)\right) $$ Thus, the denominator can be approximated by its leading term: $$ \text{Denominator} = -\frac{ex}{2} + O(x^2) $$ Step 6: Compute the limit. Substitute the simplified numerator and denominator into the limit expression for $\lambda$: $$ \lambda = \lim_{x\to 0}\frac{2x + x^2\sin\left(\frac{1}{x}\right)}{-\frac{ex}{2} + O(x^2)} $$ Divide both the numerator and the denominator by $x$: $$ \lambda = \lim_{x\to 0}\frac{2 + x\sin\left(\frac{1}{x}\right)}{-\frac{e}{2} + O(x)} $$ As $x \to 0$, the term $x\sin\left(\frac{1}{x}\right)$ approaches $0$. $$ \lambda = \frac{2 + 0}{-\frac{e}{2} + 0} = \frac{2}{-\frac{e}{2}} $$ $$ \lambda = 2 \cdot \left(-\frac{2}{e}\right) $$ $$ \lambda = -\frac{4}{e} $$