Step 1: Assume Such a Value Exists:
Suppose, for the sake of argument, some real number $\lambda$ makes $f$ continuous at $x=0$.
Then the defining condition of continuity, $\displaystyle\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)$, must hold for this $\lambda$.
Step 2: Reduce Both Sides to Numbers:
On the left branch, $x^2-2x$ can be written as $x(x-2)$, which is $0$ when $x=0$, regardless of $\lambda$.
So the left hand side of the condition becomes $\lambda\times 0$, while the right branch gives $4(0)+1=1$ on the right hand side.
The assumed condition therefore forces the equation:
\[ \lambda \times 0 = 1 \]
Step 3: Test the Equation for Contradiction:
For every real $\lambda$, multiplying by $0$ always gives $0$, never $1$.
So the equation $0=1$ is false no matter what $\lambda$ is chosen, which contradicts the assumption made in Step 1.
Since the assumption leads to a false statement, the assumption itself must be wrong.
Final Answer:
No real value of $\lambda$ can satisfy the continuity condition at $x=0$.
\[ \boxed{\text{No such } \lambda \text{ exists}} \]