Using Taylor Series to prove Limit of Exponential Function

[jemlove]

Homework Statement

Let v(λ) = ∫₀^∞ λe^-λy f(y)dy
a) Prove that limλ→ ∞ v(λ) = f(0).
b) Determine and prove limλ→ ∞ λ(v(λ)- f(0)).

α β γ δ ε ζ η θ ι κ λ μ ν ξ ο ° π ρ ς σ τ υ φ χ ψ ω Ω ~ ≈ ≠ ≡ ± ≤ ≥ Δ ∇ Σ ∂ ∫ ∏ → ∞

Homework Equations

Assume there exists some constant M where max_x∈ℝ lf(x)l < M, max_x∈ℝ lf'(x)l < M, max_x∈ℝ lf''(x)l < M.

The Attempt at a Solution

I broke up the integral into absolute values and bounded the parts. But then when I tried to find a delta I was left with four parts to the equation without the ability to isolate lambda.

 

[lippyka]

Could I get some help in finding the limit? Thanks.a) We can rewrite the integral as $$v(\lambda)=\int_0^{\infty} \lambda e^{-\lambda y}f(y)dy=\int_0^{\infty} \lambda e^{-\lambda y}(f(y)-f(0))dy+f(0).$$ Since $f(y)$ is continuous, we have that $lim_{\lambda \rightarrow \infty} \lambda e^{-\lambda y}=0$, and thus $$lim_{\lambda \rightarrow \infty}v(\lambda)=f(0).$$b) To find the limit of $\lambda (v(\lambda)-f(0))$, we can again rewrite the integral as $$\lambda (v(\lambda)-f(0))=\int_0^{\infty} \lambda^2 e^{-\lambda y}(f(y)-f(0))dy.$$ Using the same argument as before, we can see that $lim_{\lambda \rightarrow \infty} \lambda^2 e^{-\lambda y}=0$, and thus $$lim_{\lambda \rightarrow \infty}\lambda (v(\lambda)-f(0))=0.$$