# How do you evaluate #sin(x-3)/(x^2+4x-21)# as x approaches 3?

Using L-Hospital rule,i.e differentiate numerator and denominator separately without using the quotient rule, we get,

Putting x=3,

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The limit is equal to

In other words, take the derivative of the top and bottom of the fraction, then plug in the value. Let's do that:

That's the limit. We can observe this from the graph of the function:

graph{sin(x-3)/(x^2+4x-21) [-1, 7, -0.02, 0.2]}

Hope this helped!

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To evaluate sin(x-3)/(x^2+4x-21) as x approaches 3, we can substitute x=3 into the expression. However, this would result in an undefined expression since the denominator would be zero. Therefore, we need to use a different approach. By factoring the denominator, we can rewrite it as (x-3)(x+7). Now, we can cancel out the common factor of (x-3) in the numerator and denominator. This leaves us with sin(0)/(x+7). As x approaches 3, the expression sin(0)/(x+7) simplifies to 0/10, which equals 0. Therefore, the value of sin(x-3)/(x^2+4x-21) as x approaches 3 is 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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- How do you find the limit of # (sqrt(x^2+10x+1)-x)# as x approaches #oo#?
- How do you find the limit of #lnx/(sqrtx+lnx)# as #x->oo#?

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