How do you find the limit of #(x-3)/(x^2+4)# as #x->1#?

Answer 1

#lim_{x rarr 1} {x-3}/{x^2 + 4} = - 2/5#

When we have not any indeterminate form and limit exists, the value of a limit #lim_{x rarr a} f(x)# it's just #f (a)#. It means that:

#lim_{x rarr 1} {x - 3}/{x^2 + 4} = {1 - 3}/{1^2 + 4} = - 2/5#.

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Answer 2

Luke Alvarez

To find the limit of (x-3)/(x^2+4) as x approaches 1, we can substitute the value of x into the expression. Thus, we have (1-3)/(1^2+4). Simplifying this further, we get -2/5. Therefore, the limit of (x-3)/(x^2+4) as x approaches 1 is -2/5.

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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 the limit of #(x-3)/(x^2+4)# as #x-&gt;1#?

How do you find the limit of #(x-3)/(x^2+4)# as #x->1#?