What is the local linearization of #e^sin(x)# near x=1?

Answer 1

# f(x) ~~ 1.2534x+1.0664 #

Let:

# f(x) = e^(sinx) #
The linear approximation of a function #f(x)# about #x=1# is given by the linear terms of the Taylor Series about the point #x=a#, ie the terms given by:
# f(x) ~~ f(a) + f'(a)(x-a) #
Differentiating #f(x)# using the chain rule, we have:
# f'(x) = e^(sinx) (cosx) #
So with #x=1# we have:
# f(1) = e^(sin1) # # " " = 2.3197768 ... #
# f'(1) = e^(sin1) (cos1) # # " " = 1.2533897 ... #
Hence, the linear approximation near #x=1# is given by:
# f(x) ~~ 2.3198 + 1.2534(x-1) # # " " = 2.3198 + 1.2534x-1.2534 # # " " = 1.2534x+1.0664 #

Where we have rounded to 4dp.

Example:

Consider the case #x=1.01# which is near #x=1#, Then:
# f(1.01) = e^(sin(1.01)) # # " " = 2.332246 ... #

And the linear approximation gives us:

# f(1.01) ~~ 1.2534(1.01)+1.0664 # # " " = 1.265934+1.0664 # # " " = 2.332334 #
Which is correct within #3dp#
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Answer 2

The local linearization of (e^{\sin(x)}) near (x = 1) is (e^{\sin(1)} + \cos(1) \cdot (x - 1)).

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Answer from HIX Tutor

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.

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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