# How do you find the limit #lim_(x->5)(x^2-6x+5)/(x^2-25)# ?

Explanation

Using Finding Limits Algebraically,

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To find the limit of the given expression, we can directly substitute the value of x into the expression and simplify. However, if the expression is undefined at that value, we need to use algebraic manipulation to simplify it before substituting. In this case, the expression is undefined when the denominator is equal to zero.

The denominator, x^2 - 25, can be factored as (x - 5)(x + 5). Therefore, the expression becomes (x^2 - 6x + 5)/((x - 5)(x + 5)).

Now, we can cancel out the common factor of (x - 5) in the numerator and denominator.

After canceling, the expression simplifies to (x - 1)/(x + 5).

Now, we can substitute the value of x into the simplified expression.

When x approaches 5, the expression becomes (5 - 1)/(5 + 5) = 4/10 = 2/5.

Therefore, the limit of the given expression as x approaches 5 is 2/5.

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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.

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.

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.

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- What is the limit as x approaches infinity of #cos(1/x)#?
- If limit of #f(x)=2# and #g(x)=3# as #x->c#, what the limit of #5g(x)# as #x->c#?

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