How do you find the limit of #((2x^2-6)/(5x-x^2))# as x approaches infinity?
1 Answer
Explanation:
This is one way to approach this problem :
If you notice the highest degree of both the denominator and numerator is 2. We can divide every term by
#lim_(x->oo) ((2(cancelx^2)/cancel(x^2) -6/x^2) /((5x)/x^2 -cancel(x^2/x^2)))#
#lim_(x->oo)(2-6/x^2)/(5/x-1)#
#lim_(x->oo)(2-6/(oo)^2)/(5/oo-1)#
Note: As the denominator get larger, the number will be smaller, and almost close to 0. We can stated as follow
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The other method is to use L'Hopitals' Rule
If we direct substitute we will get an intermediate form
Direct Sub:
We can simply differentiate numerator and denominator separately like so
Note: Derivative of
Derivative of
#5x-x^2 = 5-2x#
We can rewrite it as :
by direct substitution, we get
We can differentiate again to get