Question
Solved Calculus Limits: Rationalization and Derivative
Original question: (d) ; (e) ;
Expert Verified Solution
Key takeaway: Both parts reward the same habit: look for a standard algebraic move first, then compare what remains with a familiar calculus pattern. One is a rationalization limit; the other is literally a derivative definition in disguise.
(d)
This is a classic rationalization problem. Multiply top and bottom by the conjugate:
Then the numerator becomes
So the expression simplifies to
Now take the limit:
(e)
Rewrite the powers as fractions:
Combine the numerator:
Cancel :
Now let :
Final answers
- (d)
- (e)
Part (e) is also the derivative of at .
Pitfalls the pros know 👇 For the radical limit, the main trap is stopping after rationalizing but forgetting that the in the denominator cancels completely. If you leave a zero in the denominator, something went wrong algebraically. For the difference quotient, don’t treat it as a raw limit of fractions before simplifying; the subtraction in the numerator has to be combined carefully before the cancels.
What if the problem changes? If the radical expression were instead of dividing by , the behavior would be different because the denominator would shrink more slowly. For part (e), if the base point changed from to another number , the same pattern would give
which is the derivative of at .
Tags: conjugate rationalization, difference quotient, derivative from first principles
FAQ
How do you solve lim (t→0) [√(t²+9)-3]/t²?
Multiply numerator and denominator by the conjugate √(t²+9)+3, simplify to 1/(√(t²+9)+3), and take the limit to get 1/6.
What is lim (h→0) [(1/(3+h) - 1/3)/h]?
Combine the fractions in the numerator, cancel h, and evaluate the limit to get -1/9.
Why is the second limit related to derivatives?
It matches the definition of the derivative of f(x) = 1/x at x=3, which is -1/x² evaluated at 3, giving -1/9.