# If the function #f(x) = (1 - x)tan(pix/2)# is continuous at #x = 1#, then #f(1)# is ??

The function

If we want the function

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If the function ( f(x) = (1 - x)\tan\left(\frac{\pi x}{2}\right) ) is continuous at ( x = 1 ), then ( f(1) ) exists and is equal to the limit of the function as ( x ) approaches 1.

To find ( f(1) ), we substitute ( x = 1 ) into the function:

[ f(1) = (1 - 1)\tan\left(\frac{\pi}{2}\right) ]

Since ( \tan\left(\frac{\pi}{2}\right) ) is undefined, we need to evaluate the limit of the function as ( x ) approaches 1 to determine if it is continuous at ( x = 1 ).

[ \lim_{x \to 1} (1 - x)\tan\left(\frac{\pi x}{2}\right) ]

To evaluate this limit, we can use the properties of limits. Since ( \tan\left(\frac{\pi}{2}\right) ) is undefined, we need to rewrite the expression.

We know that as ( x ) approaches 1, ( \frac{\pi x}{2} ) approaches ( \frac{\pi}{2} ).

Therefore, we rewrite the expression as:

[ \lim_{x \to 1} (1 - x)\tan\left(\frac{\pi}{2}\right) ]

Now, we can directly substitute ( x = 1 ) into the expression:

[ (1 - 1)\tan\left(\frac{\pi}{2}\right) = 0 ]

So, if the function ( f(x) = (1 - x)\tan\left(\frac{\pi x}{2}\right) ) is continuous at ( x = 1 ), then ( f(1) = 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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