Without using L'Hopital's rule, how can I calaculate this limit?

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SUMMARY

The limit of the function (xn - an) / (x - a) as x approaches a can be calculated by factoring the numerator. Specifically, for positive integer values of n, the expression can be factored as (x - a)(xn-1 + xn-2a + ... + an-1). This eliminates the indeterminate form and allows for direct substitution of x = a, yielding the limit as n.an-1.

PREREQUISITES
  • Understanding of polynomial functions and their limits
  • Knowledge of factoring techniques for expressions like xn - an
  • Familiarity with the concept of indeterminate forms in calculus
  • Basic understanding of limits in calculus
NEXT STEPS
  • Study polynomial limit evaluation techniques without L'Hopital's rule
  • Learn about the Factor Theorem and its applications in limits
  • Explore the concept of continuity and differentiability in relation to limits
  • Investigate higher-order derivatives and their role in limit calculations
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Students of calculus, mathematics educators, and anyone seeking to deepen their understanding of limit calculations without relying on L'Hopital's rule.

Calabi_Yau
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Without using L'Hopital's rule how can I calculate the limit of this function: (xn-an)/(x-a) when x→a

I cannot get rid of the indeterminations no matter what. I would like if you could help me out on this.
 
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Calabi_Yau said:
Without using L'Hopital's rule how can I calculate the limit of this function: (xn-an)/(x-a) when x→a

I cannot get rid of the indeterminations no matter what. I would like if you could help me out on this.

Factor x^n-a^n. For example x^2-a^2=(x-a)(x+a). x^3-a^3=(x-a)(x^2+xa+a^2) etc.
 
Calabi_Yau said:
Without using L'Hopital's rule how can I calculate the limit of this function: (xn-an)/(x-a) when x→a

I cannot get rid of the indeterminations no matter what. I would like if you could help me out on this.

Is n a positive integer? If so, just factor, as Dick has already suggested. If n is not a positive integer, you more-or-less need to use l'Hosptial's rule, whether you want to or not.
 
I figured that out. Took me a while to notice if I factorized x^n-a^n beginning with (x-a)(...) the sum of the other factors would add up to n.a^n-1.

PS: I apologize not having posted my original atempt in solving the problem like the rules require.
 

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