Determining a trigonometric limit

AI Thread Summary
To calculate the trigonometric limit, users suggested various approaches, including multiplying by the conjugate and substituting variables. It was noted that the function is continuous at x=0, allowing for direct substitution. If the denominator involves cube roots, L'Hospital's rule may be applicable. Additionally, an algebraic identity for cubic differences could be useful if calculus methods are not preferred. The discussion emphasizes exploring multiple algebraic techniques to solve the limit problem effectively.
Bunny-chan
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Homework Statement


Calculate the following limit:

png.png


Homework Equations

The Attempt at a Solution


I don't know how to proceed with this. I've tried to multiply by the conjugate, and to simplify the expression (x+\pi) to u, but I wasn't very sucessful. To what kind of algebric device I could resort to? Or is there other way to deduce the limit?
 
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Bunny-chan said:

Homework Statement


Calculate the following limit:

View attachment 203337

Homework Equations

The Attempt at a Solution


I don't know how to proceed with this. I've tried to multiply by the conjugate, and to simplify the expression (x+\pi) to u, but I wasn't very sucessful. To what kind of algebric device I could resort to? Or is there other way to deduce the limit?
The problem as written is continuous at ##x=0##, so just plug it in. Or if both terms in the denominator are supposed to be cube roots, try L'Hospital's rule.
 
Bunny-chan said:

Homework Statement


Calculate the following limit:

View attachment 203337

Homework Equations

The Attempt at a Solution


I don't know how to proceed with this. I've tried to multiply by the conjugate, and to simplify the expression (x+\pi) to u, but I wasn't very sucessful. To what kind of algebric device I could resort to? Or is there other way to deduce the limit?
I assume you meant to write ##\sqrt[3]{x+\pi}## instead of ##3 \sqrt{x+\pi}## in the denominator. If you do not want to (or are unable to) use calculus, use instead the algebraic identity ##a^3-b^3 = (a-b)(a^2+a b + b^2)## for appropriate ##a## and ##b##.
 
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