Some limit problems

mathdaemon

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Hello

Find the limit of √(4-x²) as x approaches 2 ?

Find the limit of tan x as x approaches
Π/2 ?

Answer for both is "does not exist".

My solution :
√(4-2²) = 0 and tan Π/2=

These look like such simple problems. I can't believe I am asking you this. Where am I going wrong ?

Thank you


 
Hello

Find the limit of √(4-x²) as x approaches 2 ?

Find the limit of tan x as x approaches
Π/2 ?

Answer for both is "does not exist".

My solution :
√(4-2²) = 0 and tan Π/2=

These look like such simple problems. I can't believe I am asking you this. Where am I going wrong ?

Thank you


For the tangent function, since infinity is not a real number and, at this time, you are considering only real numbers, the limit does not exist.

For the \(\displaystyle \sqrt{4 - x^2}\), it is a little trickier. When we say the limit exists, we mean for any epsilon greater than zero when ever |x-2| is less than delta then ... OK, suppose there were a limit and we have the delta for a given epison. Now let x = 2+\(\displaystyle \frac{\delta}{2}\). What is \(\displaystyle \sqrt{4 - x^2}\)? Well
\(\displaystyle 4 - x^2 = 4 - (2 + \frac{\delta}{2})^2 = 4 - (4 + 2 \frac{\delta}{2} + (\frac{\delta}{2})^2 = -(2 \frac{\delta}{2} + (\frac{\delta}{2})^2)\)
and we have the square root of a negative number.
 
Hello

tan(x) = sin(x)/cos(x). Then, for 0 < x < π/2, cos(x) > 0 and sin(x) > 0. This implies that sin(x) --> 1 and cos(x) --> 0+.

Therefore, the limit is +infinity.

Similarly for
π/2<x< π ,sin(x) >0 and cos(x)<0. Therefore the limit is -infinity.

Hope I have got it correct.

Please Please reply
 
Hello

tan(x) = sin(x)/cos(x). Then, for 0 < x < π/2, cos(x) > 0 and sin(x) > 0. This implies that sin(x) --> 1 and cos(x) --> 0+.

Therefore, the limit is +infinity.

Similarly for
π/2<x< π ,sin(x) >0 and cos(x)<0. Therefore the limit is -infinity.

Hope I have got it correct.

Please Please reply
Basically yes you did get it correct.

However, when you said "This implies that sin(x) --> 1 and cos(x) --> 0+. Therefore, the limit is +infinity. ", the proper English would be "Since sin(x) --> 1 and cos(x) --> 0+ as x-->π/2-, this implies the limit is +infinity as x-->π/2-". For the other side, it would be "... Therefore the limit is -infinity as x-->π/2+."
 
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