Let f: R → R be defined as f(x)= {x^5 sin (1/x)+5 x^2, x<0; 0, x=0; x^5 cos (1/x)+λ x^2, x>0} The value of λ for which f^′ ′(0) exists, is \ \ \ \.

Let $\displaystyle f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as $$f(x)=\left\{\begin{array}{cc} x^5 \sin \left(\frac{1}{x}\right)+5 x^2, & x<0 \\ 0, & x=0 \\ x^5 \cos \left(\frac{1}{x}\right)+\lambda x^2, & x>0 \end{array}\right. $$ The value of $\displaystyle \lambda$ for which $\displaystyle f^{\prime \prime}(0)$ exists, is \_\_\_\_.