Respuesta :

LammettHash
You can recognize the limit as a derivative. Recall that

[tex]f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h[/tex]

and more to the point,

[tex]f'(1)=\displaystyle\lim_{h\to0}\frac{f(1+h)-f(1)}h[/tex]

We have

[tex]f(x)=x^4\implies f'(x)=4x^3\implies f'(1)=4[/tex]

so the answer is B.

We don't actually have to invoke the definition of the derivative. Instead we can just use the definition of [tex]f(x)[/tex]:

[tex]\displaystyle\lim_{h\to0}\frac{(1+h)^4-1^4}h=\lim_{h\to0}\frac{1+4h+6h^2+4h^3+h^4-1}h=\lim_{h\to0}(4+6h+4h^2+h^3)[/tex]

which you can see also approaches 4.

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