Math  /  Calculus

QuestionIn this problem, we will consider ff to be some function defined on a subset of the real numbers, and we'll assume all of its derivative exist everywhere. We will consider the limit L=limx5f(5)+f(2x5)f(5ex24x5)1L=\lim _{x \rightarrow 5} \frac{f^{\prime}(5)+f^{\prime \prime}(2 x-5)}{f\left(5 e^{x^{2}-4 x-5}\right)-1}
For each of parts (a) and (b), you will determine LL if it is a real number. Enter inf or -inf if the limit is ±\pm \infty, or enter dne if the limit fails to exist in a different way. (a) Suppose that the fourth-order Taylor polynomial at x=5x=5 is equal to T3(x)=4(x5)2(x5)2+2x5)3+100(x5)4\left.T_{3}(x)=4(x-5)-2(x-5)^{2}+2 x-5\right)^{3}+100(x-5)^{4}
Then L=L= \square (b) Suppose that the degree 4 Taylor polynomial at x=5x=5 is equal to T3(x)=1+4(x5)2(x5)2+2(x5)3+100(x5)4T_{3}(x)=1+4(x-5)-2(x-5)^{2}+2(x-5)^{3}+100(x-5)^{4}
Then L=L= \square

Studdy Solution
(a) L=0L = 0 (b) L=L = dne

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