For α,β,γ∈R, if x→0limsin(2x)−βxx2sin(αx)+(γ−1)ex2=3, then β+γ−α is equal to:
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Step-by-Step Solution
Step 1: Analyze the Denominator
First, let's analyze the denominator of the given limit. For the limit to be finite and non-zero, the denominator must approach zero as x→0. We use the Taylor series expansion for sin(2x) around x=0, which is sin(u)=u−3u3!+…. Substituting u=2x, we get the expansion for sin(2x). Then we combine it with −βx.
Step 2: Determine the value of β
If 2−β=0, the denominator would be of order x, while the numerator (as we will see) would be of order x2 or higher, making the limit 0 or ∞, not 3. Therefore, for the limit to be finite and non-zero, the coefficient of x in the denominator must be zero. This implies 2−β=0, so β=2.
Step 3: Analyze the Numerator
Now, let's analyze the numerator. We use the Taylor series expansions for sin(αx) and ex2 around x=0. The expansion for sin(u) is u−3u3!+…, so sin(αx)=αx−3(αx)3!+…. The expansion for eu is 1+u+2u2!+…, so ex2=1+x2+2(x2)2!+…. We substitute these into the numerator and collect terms.
Step 4: Determine the value of γ
For the limit to be finite and non-zero, the lowest power of x in the numerator must be at least x3 (since the denominator is of order x3 after setting β=2). This means the constant term and the x2 term in the numerator must be zero. Thus, γ−1=0, which implies γ=1.
Step 5: Evaluate the Limit and Determine α
With β=2 and γ=1, the limit simplifies to x→0lim−34x3+O(x5)αx3+O(x4). We can cancel x3 from the numerator and denominator, and then evaluate the limit by taking the ratio of the coefficients of x3. We set this equal to 3 and solve for α.
Step 6: Calculate β+γ−α
Finally, we substitute the values of α, β, and γ that we found into the expression β+γ−α to get the final answer.