Q. If L'Hospital's Rule applies, use it to evaluate the limit.(Use symbolic notation and fractions where needed.)limx→−29725−4x−x2x2−841=
Check Indeterminate Form: First, we need to determine if L'Hôpital's Rule can be applied. L'Hôpital's Rule can be used when the limit results in an indeterminate form like 0/0 or ∞/∞. Let's evaluate the function at x=−29 to see if we get an indeterminate form.
Evaluate at x=−29: Substitute x=−29 into the numerator and denominator:Numerator: (−29)2−841=841−841=0Denominator: 725−4(−29)−(−29)2=725+116−841=0Since both the numerator and denominator evaluate to 0, we have an indeterminate form of 0/0, and L'Hôpital's Rule can be applied.
Apply L'Hôpital's Rule: Now we will apply L'Hôpital's Rule, which tells us to take the derivative of the numerator and the derivative of the denominator and then evaluate the limit of the new function.
Find Derivatives: The derivative of the numerator with respect to x is: dxd(x2−841)=2x. The derivative of the denominator with respect to x is: dxd(725−4x−x2)=−4−2x.
Evaluate New Limit: Now we have a new limit to evaluate: limx→−29−4−2x2x
Substitute x=−29: Substitute x=−29 into the new function:(2×−29)/(−4−2×−29)=(−58)/(−4+58)=−58/54
Simplify Fraction: Simplify the fraction−5458 by dividing both the numerator and the denominator by their greatest common divisor, which is 2:54−58=27−29
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