Let the numerator of our function be called f (x), and let the denominator be called...

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Let the numerator of our function be called f (x), and let the denominator be called g(x). That is, • f (x) = sin x and g(x)=x²-7². Important: no quotient rule f(x) is a quotient, we will not be using the quotient rule in this problem. The quotient rule g(x) f(x) would find the derivative of h (x) = (2). However, we want to find the limit of h (x) as a approaches , not g(x) the derivative. Although h (x) = Local linearization Again, consider f (x) = sin x. Find the local linearization of f (x) at x = πT. Call this tangent line equation Lf (x). I suggest the notation Lfas an abbreviation for the "L"inearization of function "f." Hints: • You'll need to find f' (π), and you'll need to find the y-value that corresponds to x = π. • Next, you can plug those pieces of information into the point-slope formula, Lf (x) - y₁ = m (x − x₁). Then, rearrange and simplify. ⒸL₁ (x) = -x + T ⒸL₁ (x) = -x - 2π Lf (x) = -2x + π Let the numerator of our function be called f (x), and let the denominator be called g(x). That is, • f (x) = sin x and g(x)=x²-7². Important: no quotient rule f(x) is a quotient, we will not be using the quotient rule in this problem. The quotient rule g(x) f(x) would find the derivative of h (x) = (2). However, we want to find the limit of h (x) as a approaches , not g(x) the derivative. Although h (x) = Local linearization Again, consider f (x) = sin x. Find the local linearization of f (x) at x = πT. Call this tangent line equation Lf (x). I suggest the notation Lfas an abbreviation for the "L"inearization of function "f." Hints: • You'll need to find f' (π), and you'll need to find the y-value that corresponds to x = π. • Next, you can plug those pieces of information into the point-slope formula, Lf (x) - y₁ = m (x − x₁). Then, rearrange and simplify. ⒸL₁ (x) = -x + T ⒸL₁ (x) = -x - 2π Lf (x) = -2x + π

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