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Given that the limit as 饾懃 approaches negative one of 饾懃 squared minus 饾憵 minus one times 饾懃 minus 饾憵 all divided by 饾懃 plus one is equal to negative three, determine the value of 饾憵.
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The question tells us that the limit as 饾懃 approaches negative one of a rational function is equal to negative three.
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It wants us to use this information to determine the value of 饾憵.
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Since this is the limit of a rational function, we can attempt to evaluate this by using direct substitution.
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Substituting 饾懃 is equal to negative one in the function in our limit gives us negative one squared minus 饾憵 minus one times negative one minus 饾憵 all divided by negative one plus one.
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Fully evaluating the expressions in our numerator and our denominator, we see we get the indeterminate form zero divided by zero.
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So evaluating our limit by direct substitution gained us no information about the value of 饾憵.
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So we鈥檙e going to need a different way to evaluate this limit.
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We recall that L鈥橦么pital鈥檚 rule gives us a way of evaluating an indeterminate limit of this form.
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We recall the following version of L鈥橦么pital鈥檚 rule.
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If we have two differentiable functions 饾憮 and 饾憯 where 饾憯 prime of 饾懃 is not equal to zero around our value of 饾憥 except possibly when 饾懃 is equal to 饾憥, and the limit as 饾懃 approaches 饾憥 of 饾憮 of 饾懃 and the limit as 饾懃 approaches 饾憥 of 饾憯 of 饾懃 are both equal to zero.
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Then the limit as 饾懃 approaches 饾憥 of the quotient of 饾憮 of 饾懃 and 饾憯 of 饾懃 is equal to the limit as 饾懃 approaches 饾憥 of the quotient of 饾憮 prime of 饾懃 and 饾憯 prime of 饾懃.
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In this version of L鈥橦么pital鈥檚 rule, we鈥檙e told that if the limit as 饾懃 approaches 饾憥 of 饾憮 of 饾懃 over 饾憯 of 饾懃 gives us the indeterminate form of zero over zero.
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Then, under these conditions, we can instead calculate the limit as 饾懃 approaches 饾憥 of 饾憮 prime of 饾懃 divided by 饾憯 prime of 饾懃.
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Let鈥檚 start by verifying that we can actually use L鈥橦么pital鈥檚 rule in this case.
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We鈥檒l set our function 饾憮 of 饾懃 equal to the numerator of the function in our limit.
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That鈥檚 饾懃 squared minus 饾憵 minus one times 饾懃 minus 饾憵.
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And we鈥檒l set our function 饾憯 of 饾懃 to be the function in the denominator of our limit.
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That鈥檚 饾憯 of 饾懃 is equal to 饾懃 plus one.
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The first thing we need to use this version of L鈥橦么pital鈥檚 rule is that both our functions 饾憮 and 饾憯 are differentiable.
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We see the 饾憯 is a linear function, and for any value 饾憵, 饾憮 of 饾懃 is a quadratic.
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So both of these functions are polynomials, and therefore they鈥檙e both differentiable.
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Next, we need to show that derivative of 饾憯 of 饾懃 is not equal to zero around 饾憥.
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Remember, 饾憥 is the constant that 饾懃 is approaching in our limit.
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We have the limit as 饾懃 approaches negative one.
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So we鈥檒l set 饾憥 equal to negative one.
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Since 饾憯 of 饾懃 is the linear function 饾懃 plus one, we can just calculate 饾憯 prime of 饾懃 directly by using the power rule for differentiation.
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We multiply by the exponent and reduce the exponent by one.
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In this case, we just get the constant one.
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And we can, of course, see that this is never equal zero.
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Next, we need to show that the limit as 饾懃 approaches negative one of 饾憮 of 饾懃 and the limit as 饾懃 approaches negative one of 饾憯 of 饾懃 is both equal to zero.
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However, we鈥檝e actually already shown this.
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When we tried to use direct substitution to evaluate our limit, we saw that our numerator evaluated to give us zero and our denominator also evaluated to give us zero.
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But the numerator was just the limit as 饾懃 approaches negative one of 饾憮 of 饾懃.
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And the denominator was the limit as 饾懃 approaches negative one of 饾憯 of 饾懃.
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So we鈥檝e shown the last two prerequisites of this version of the L鈥橦么pital鈥檚 rule is true.
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So we can attempt to use it to evaluate our limit.
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So by using this version of L鈥橦么pital鈥檚 rule, we have the limit given to us in the question, which is the limit as 饾懃 approaches negative one of 饾憮 of 饾懃 divided by 饾憯 of 饾懃 is equal to the limit as 饾懃 approaches negative one of 饾憮 prime of 饾懃 divided by 饾憯 prime of 饾懃.
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So to evaluate this limit, we need to find 饾憮 prime of 饾懃.
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We recall 饾憮 of 饾懃 is equal to 饾懃 squared minus 饾憵 minus one times 饾懃 minus 饾憵.
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And for any value of 饾憵, this is just a quadratic.
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So we can evaluate this derivative by using the power rule for differentiation.
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We get two 饾懃 minus 饾憵 minus one.
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So now that we found 饾憮 prime of 饾懃 and 饾憯 prime of 饾懃, we鈥檝e shown that our limit is equal to the limit as 饾懃 approaches negative one of two 饾懃 minus 饾憵 minus one divided by one.
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Of course, dividing by one inside of our limit doesn鈥檛 change anything.
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And now we see we鈥檙e just calculating the limit of a linear function.
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We can evaluate this by using direct substitution.
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We substitute 饾懃 is equal to negative one into our linear function to evaluate the limit.
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And substituting 饾懃 is equal to negative one into this limit gives us two times negative one minus 饾憵 minus one.
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And if we evaluate and simplify this expression, we can see that it鈥檚 equal to negative one minus 饾憵.
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So we鈥檝e shown that the limit given to us in the question is equal to negative one minus 饾憵.
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But the question also tells us that this limit must be equal to negative three.
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So, negative one minus 饾憵 is equal to negative three.
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And we can then solve this equation.
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We get that 饾憵 is equal to two.
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So we鈥檝e shown if the limit as 饾懃 approaches negative one of 饾懃 squared minus 饾憵 minus one times 饾懃 minus 饾憵 all divided by 饾懃 plus one is equal to negative three, then the value of 饾憵 must be equal to two.