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The limit `lim_(h->0) (3(x+h)^47 -(3x^47))/h` has to be determined.
If h is equated to 0 the form `0/0` is obtained which is indeterminate. This allows the use of l'Hopital's rule and the numerator and denominator are substituted by their derivatives.
Here as h tends to 0, the derivative is with respect to h
=> `lim_(h->0) (3*47*(x+h)^46 - 0)/1`
substitute h = 0
The required limit is 141*x^46
Evaluate lim as h approaches 0 (3(x+h)^47 -(3X^47))/h
`lim_(h->0) ( 3(x+h)^47 - (3x^47) )/ (h)`
``There is a much simpler way to solve this than using L'Hopital's rule. It requires you to realize that this limit is in the form:
`lim_(h->0) ( f(x+h) - f(x) ) / (h)`
This is the formal definition of the derivative. All you have to do is take the derivative of f(x)!
In your case, f(x) = 3x^47.
Therefore f'(x) = 141x^46
and that is what the limit is equal to.
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