Evaluate the Limit (just confirmation)

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SUMMARY

The limit of the expression (cubicsquareroot(x) - 2)/(x-8) as x approaches 8 evaluates to 1/12. The solution involves substituting p for cubicsquareroot(x), leading to the limit form of (p - 2)/(p^3 - 8). The denominator can be factored into (p - 2)(p^2 + 2p + 4), allowing for the limit to be calculated without the need for L'Hôpital's Rule. The final answer is confirmed as correct by multiple participants in the discussion.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with cubed roots and their properties
  • Knowledge of factoring polynomials
  • Basic understanding of L'Hôpital's Rule
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  • Study the application of L'Hôpital's Rule in limit problems
  • Learn about polynomial factoring techniques
  • Explore the properties of cubed roots and their derivatives
  • Practice solving limits involving indeterminate forms
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Students studying calculus, particularly those focusing on limits and indeterminate forms, as well as educators looking for examples of limit evaluation techniques.

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Homework Statement



lim (cubicsquareroot(x) - 2)/(x-8)
x→8



Homework Equations





The Attempt at a Solution



I obtained 1/12 as my final answer, after changing the variable since its in indeterminate form.
..let p=cubicsquareroot(x), then as x→8, p→2
then, p^3=x

Using this logic I obtained 1/12.

Is this correct ?
 
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NATURE.M said:

Homework Statement



lim (cubicsquareroot(x) - 2)/(x-8)
x→8



Homework Equations





The Attempt at a Solution



I obtained 1/12 as my final answer, after changing the variable since its in indeterminate form.
..let p=cubicsquareroot(x), then as x→8, p→2
then, p^3=x

Using this logic I obtained 1/12.

Is this correct ?

Precisely, Although It'd be nicer if you had done some substitutions.. Yeah is correct!
 
PrashntS said:
Precisely, Although It'd be nicer if you had done some substitutions.. Yeah is correct!

My bad! you have! And when you reach L'Hopital's rule, revisit these questions..
 
What's a "cubicsquareroot" ?
 
SammyS said:
What's a "cubicsquareroot" ?

Assumed he meant x^(1/3) *kids today*
 
lol Yeah I mean't x^(1/3). And thanks.
 
PrashntS said:
My bad! you have! And when you reach L'Hopital's rule, revisit these questions..
L' Hopital's Rule is not needed in this problem.

The OP apparently revised the original problem to this form:
$$ \lim_{p \to 2} \frac{p - 2}{p^3 - 8}$$

and then factored the denominator to (p - 2)(p2 + 2p + 4), and then took the limit.
 

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