How do you show that the linearization of #f(x) = (1+x)^k# at x=0 is #L(x) = 1+kx#?
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To show that the linearization of ( f(x) = (1+x)^k ) at ( x = 0 ) is ( L(x) = 1 + kx ), you can use the definition of linearization. The linearization of a function ( f(x) ) at a point ( x = a ) is given by:
[ L(x) = f(a) + f'(a)(x  a) ]

First, find the value of ( f(0) ) by substituting ( x = 0 ) into the function ( f(x) ). [ f(0) = (1+0)^k = 1^k = 1 ]

Next, find the derivative of ( f(x) ) with respect to ( x ), ( f'(x) ). [ f'(x) = k(1+x)^{k1} ]

Evaluate ( f'(0) ) by substituting ( x = 0 ) into ( f'(x) ). [ f'(0) = k(1+0)^{k1} = k ]

Plug ( f(0) ) and ( f'(0) ) into the formula for linearization at ( x = 0 ). [ L(x) = f(0) + f'(0)(x  0) ] [ L(x) = 1 + kx ]
Thus, the linearization of ( f(x) = (1+x)^k ) at ( x = 0 ) is indeed ( L(x) = 1 + kx ).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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