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How do you find #lim (6x^3-4)/(2x+5)# as #x->oo#?

Answer 1

You should get #oo#

Consider the numerator first: even if you subtract #4# as #x# increases it will become very big and rapidly. The denominator as well BUT not as fast as the numerator! So the numerator will "win"and in the limit the function will tend to #oo#.

We can also write collecting #x#: #lim_(x->oo)(cancel(x)(6x^2-4/x))/(cancel(x)(2+5/x))# as #x->oo# #4/x->0# and #5/x->0# giving: #lim_(x->oo)(cancel(x)(6x^2-4/x))/(cancel(x)(2+5/x))=oo#

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Answer 2

Luke Alvarez

To find the limit of (6x^3-4)/(2x+5) as x approaches infinity, we can use the concept of limits. By dividing each term in the numerator and denominator by the highest power of x, we can simplify the expression. In this case, we divide both the numerator and denominator by x^3. This yields (6-4/x^3)/(2/x+5/x^3). As x approaches infinity, the terms with 1/x^3 become negligible, resulting in (6-0)/(0+0). Simplifying further, we get 6/0, which is undefined. Therefore, the limit of (6x^3-4)/(2x+5) as x approaches infinity does not exist.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find #lim (6x^3-4)/(2x+5)# as #x-&gt;oo#?

How do you find #lim (6x^3-4)/(2x+5)# as #x->oo#?