# How do you show the limit does not exist #lim_(x->4)(x-4)/(x^2-8x+16)# ?

Solution

Now applying the limit, we get

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To show that the limit does not exist, we need to find two different paths approaching the limit point where the function gives different values. In this case, we can consider approaching the limit from the left and right sides of x=4.

Approaching from the left side (x<4): Substituting x=4-h into the function, where h is a positive number approaching zero, we get: lim_(h->0) ((4-h)-4)/((4-h)^2-8(4-h)+16) Simplifying this expression, we have: lim_(h->0) (-h)/(h^2-8h+16) Taking the limit as h approaches zero, we get: lim_(h->0) (-h)/(h^2-8h+16) = 0/16 = 0

Approaching from the right side (x>4): Substituting x=4+h into the function, where h is a positive number approaching zero, we get: lim_(h->0) ((4+h)-4)/((4+h)^2-8(4+h)+16) Simplifying this expression, we have: lim_(h->0) (h)/(h^2+8h+16) Taking the limit as h approaches zero, we get: lim_(h->0) (h)/(h^2+8h+16) = 0/16 = 0

Since the limit from both the left and right sides is 0, we can conclude that the limit as x approaches 4 does exist and is equal to 0.

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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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