Solving the Limit of sqrt(x^2-9)/abs(3-x)

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In summary, the limit of sqrt(x^2-9)/abs(3-x) as x approaches 3 is undefined due to the denominator becoming zero. It can be solved using L'Hopital's rule or by factoring and simplifying. The absolute value in the denominator ensures that the expression is defined for all values of x, including when x equals 3. However, the limit may be different for positive and negative values of x. The limit cannot be evaluated at x=3 because the expression is undefined at that point.
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gipc
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I'm sorry for being pathetic but I can't seem to solve anything this days.

[tex]lim sqrt(x^2-9)/abs(3-x) as x->3+[/tex]
 
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Hi gipc! :smile:

You mean limx->3+ √(x2 - 9) / |3 - x| ?

Well x -> 3+ means you only have to consider x > 3, so |3 - x| = x - 3 …

does that help? :smile:​
 

FAQ: Solving the Limit of sqrt(x^2-9)/abs(3-x)

What is the limit of sqrt(x^2-9)/abs(3-x) as x approaches 3?

The limit of sqrt(x^2-9)/abs(3-x) as x approaches 3 is undefined because the denominator becomes zero.

How can the limit of sqrt(x^2-9)/abs(3-x) be solved?

The limit of sqrt(x^2-9)/abs(3-x) can be solved using L'Hopital's rule or by factoring and simplifying the expression.

What is the significance of the absolute value in the denominator of the expression sqrt(x^2-9)/abs(3-x)?

The absolute value in the denominator ensures that the expression is defined for all values of x, including when x equals 3. Without the absolute value, the expression would be undefined at x=3 due to division by zero.

Is the limit of sqrt(x^2-9)/abs(3-x) the same for both positive and negative values of x?

No, the limit may be different for positive and negative values of x. For x values approaching 3 from the positive side, the limit is 3/0, which is undefined. For x values approaching 3 from the negative side, the limit is -3/0, which is also undefined.

Can the limit of sqrt(x^2-9)/abs(3-x) be evaluated at x=3?

No, the limit cannot be evaluated at x=3 because the expression is undefined at that point.

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