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[Miwaa]

lim x----2pi sin x

 

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To find the limit of sin(x) as x approaches 2π, we can use the fact that sin(x) is a continuous function.

When x approaches 2π from the left side, sin(x) approaches 0 because sin(2π) = 0.

Similarly, when x approaches 2π from the right side, sin(x) also approaches 0.

Therefore, the limit of sin(x) as x approaches 2π is 0.

 

[Miwaa]

To find the limit of sin(x) as x approaches 2π, we can use the fact that sin(x) is a continuous function.

When x approaches 2π from the left side, sin(x) approaches 0 because sin(2π) = 0.

Similarly, when x approaches 2π from the right side, sin(x) also approaches 0.

Therefore, the limit of sin(x) as x approaches 2π is 0.

evaluate limx---3 4^x+5

 

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evaluate limx---3 4^x+5

To evaluate the limit as x approaches 3 of 4^(x+5), we can substitute the value of x into the expression.

When x approaches 3, we have:

4^(3+5) = 4^8 = 65,536

Therefore, the limit of 4^(x+5) as x approaches 3 is 65,536.

 

[Miwaa]

To evaluate the limit as x approaches 3 of 4^(x+5), we can substitute the value of x into the expression.

When x approaches 3, we have:

4^(3+5) = 4^8 = 65,536

Therefore, the limit of 4^(x+5) as x approaches 3 is 65,536.

no

valuate limx---3 4^x+5 not

3 of 4^(x+5)

 

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no

valuate limx---3 4^x+5 not

Apologies for the confusion. Let's reevaluate the limit.

To evaluate the limit as x approaches 3 of 4^(x+5), we can rewrite it as 4^x * 4^5.

As x approaches 3, 4^x approaches 4^3 = 64, and 4^5 = 1024.

Hence, the limit of 4^(x+5) as x approaches 3 is 64 * 1024 = 65536.

 

[Miwaa]

Apologies for the confusion. Let's reevaluate the limit.

To evaluate the limit as x approaches 3 of 4^(x+5), we can rewrite it as 4^x * 4^5.

As x approaches 3, 4^x approaches 4^3 = 64, and 4^5 = 1024.

Hence, the limit of 4^(x+5) as x approaches 3 is 64 * 1024 = 65536.

EValuate limx---3 4^x + 5

 

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valuate limx---3 4^x + 5

I apologize for the confusion in my previous responses. It seems I misunderstood your expression.

To evaluate the limit as x approaches 3 of 4^x + 5, we can substitute the value of x into the expression.

When x approaches 3, we have:

4^3 + 5 = 64 + 5 = 69

Therefore, the limit of 4^x + 5 as x approaches 3 is 69.

 

[Miwaa]

I apologize for the confusion in my previous responses. It seems I misunderstood your expression.

To evaluate the limit as x approaches 3 of 4^x + 5, we can substitute the value of x into the expression.

When x approaches 3, we have:

4^3 + 5 = 64 + 5 = 69

Therefore, the limit of 4^x + 5 as x approaches 3 is 69.

evaluate lim x---- -2 (2x+4)(x-5)

 

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evaluate lim x---- -2 (2x+4)(x-5)

To evaluate the limit as x approaches -2 of -2(2x+4)(x-5), we can substitute the value of x into the expression.

When x approaches -2, we have:

-2(2(-2)+4)(-2-5)
= -2(0)(-7)
= 0

Therefore, the limit of -2(2x+4)(x-5) as x approaches -2 is 0.

 

[Miwaa]

To evaluate the limit as x approaches -2 of -2(2x+4)(x-5), we can substitute the value of x into the expression.

When x approaches -2, we have:

-2(2(-2)+4)(-2-5)
= -2(0)(-7)
= 0

Therefore, the limit of -2(2x+4)(x-5) as x approaches -2 is 0.

evaluate limx---3 (x^2 +x-12/x-3)