Q. Mandisa's teacher gave her a flow chart (below) and asked her to find limx→−1f(x) for f(x)=x2−11+5x+30.Calculating limx→af(x)
Check for Undefined: First, substitute x=−1 into the function to check if it's defined. f(−1)=(−1)2−11+5(−1)+30=1−11+25=01+5=06Since the denominator is 0, the function is undefined at x=−1.
Factor Denominator: Next, factor the denominator to see if we can simplify the function.x2−1=(x−1)(x+1)So, f(x)=(x−1)(x+1)1+5x+30
Apply L'Hôpital's Rule: Now, let's use the limit properties and L'Hôpital's Rule since we have a 00 form.x→−1lim(x−1)(x+1)1+5x+30Differentiate the numerator and the denominator separately.Numerator: dxd(1+5x+30)=dxd(1)+dxd(5x+30)=0+25x+305=25x+305Denominator:dxd((x−1)(x+1))=dxd(x2−1)=2xSo, x→−1lim2x25x+305=x→−1lim4x5x+305
Simplify Limit: Substitute x=−1 into the simplified limit.x→−1lim4x5x+305=4(−1)5(−1)+305=−4255=−4⋅55=−205=−41
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