What is the equation of the line tangent to # f(x)=x^2/e^x-x/e^(x^2) # at # x=0#?

Answer 1

#f'(x)=-1#

Let's rewrite the function as #f(x)=x^2e^{-x}-xe^{-x^2}#

Reminding that the Mac Laurin of #e^x# is #1+x+x^2/2+o(x^2)#, the one of #e^{-x}# is #1-x+x^2/2+o(x^2)#, and the one of #e^{-x^2}# is # 1-x^2+x^4/2+o(x^4)# By replacing them, we get #f(x)=x^2(1-x+x^2/2+o(x^2))-x(1-x^2+x^4/2+o(x^4))#. By carrying out the multiplication we get #-x+x^2+x^4/2-x^5/2+o(x^4)# In the end #f'(0)=-1#, i.e. the coefficient of the linear term in the Mac Laurin polynomial for #f(x)# Enjoy!

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Answer 2

Scarlett Allison

The equation of the line tangent to f(x) at x=0 is y = -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.