Solving Trigonometric Limit w/o L'Hospital Rule

In summary, the conversation is about finding the limit of $\displaystyle \frac{\sin(x^3-1)}{x-1}$ as $x$ approaches 1, without using L'Hôpital's rule. The suggested method is to multiply both the numerator and denominator by $x^2 + x + 1$, which leads to finding the limit of $\displaystyle \frac{\sin(x^3-1)}{x^3-1}$. L'Hôpital's rule is also mentioned as a possible solution.
  • #1
Yankel
395
0
Hello,

I need some help with this limit, I have no clue how to do this with the next constraint: No use of L'hopital rule...

Thank you !

[tex]\lim_{x\to1}\frac{sin(x^{3}-1)}{x-1}[/tex]
 
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  • #2
Yankel said:
Hello,

I need some help with this limit, I have no clue how to do this with the next constraint: No use of L'hopital rule...

Thank you !

[tex]\lim_{x\to1}\frac{sin(x^{3}-1)}{x-1}[/tex]

Try multiplying numerator and denominator by $x^{2} + x + 1$...

Kind regards

$\chi$ $\sigma$
 
  • #3
chisigma said:
Try multiplying numerator and denominator by $x^{2} + x + 1$...

Kind regards

$\chi$ $\sigma$

I thought about it, it gives me what's in the sine, however x->1 and not x->0, so I am still stuck...
 
  • #4
Yankel said:
I thought about it, it gives me what's in the sine, however x->1 and not x->0, so I am still stuck...

$\displaystyle \lim_{x \rightarrow 1} \frac{\sin (x^{3}-1)}{x-1} = \lim_{x \rightarrow 1} \frac{\sin (x^{3}-1)}{x^{3}-1}\ (x^{2} + x + 1)$

... and now what is $\displaystyle \lim_{x \rightarrow 1} \frac{\sin (x^{3}-1)}{x^{3}-1}$?...

... and what is $\displaystyle \lim_{x \rightarrow 1} x^{2} + x + 1$?...

Kind regards

$\chi$ $\sigma$
 
  • #5
Also, L'Hôpital's rule allows us to write:

$\displaystyle \lim_{x\to1}\frac{\sin(x^3-1)}{x-1}=3\lim_{x\to1}x^2\cos(x^3-1)$
 

FAQ: Solving Trigonometric Limit w/o L'Hospital Rule

What is the L'Hospital Rule and when is it used?

The L'Hospital Rule is a mathematical technique used to solve indeterminate forms, or limits that cannot be evaluated directly. It involves taking the derivative of the numerator and denominator of a fraction and then evaluating the limit again. This rule is used when other methods, such as factoring or substitution, do not work to solve a limit.

Can trigonometric limits be solved without using the L'Hospital Rule?

Yes, trigonometric limits can be solved without using the L'Hospital Rule. Other methods, such as trigonometric identities, substitution, and factoring, can be used to solve these types of limits. However, the L'Hospital Rule may be a quicker and more efficient method in some cases.

What are some common trigonometric identities used to solve limits?

Some common trigonometric identities used to solve limits include:
- sin^2(x) + cos^2(x) = 1
- tan(x) = sin(x)/cos(x)
- cot(x) = cos(x)/sin(x)
- sec(x) = 1/cos(x)
- csc(x) = 1/sin(x)
These identities can be used to simplify trigonometric expressions and solve limits without using the L'Hospital Rule.

How can substitution be used to solve trigonometric limits?

Substitution involves replacing a variable in an expression with a specific value or function. In trigonometric limits, substitution can be used by setting the limit equal to a specific value, such as 0 or infinity, and then substituting in trigonometric identities to simplify the expression. This can help to solve the limit without using the L'Hospital Rule.

Are there any special cases when solving trigonometric limits?

Yes, there are some special cases when solving trigonometric limits. These include:
- Limits involving infinity: In these cases, the limit can be evaluated by looking at the highest degree term in the numerator and denominator.
- Limits involving trigonometric functions raised to a power: These limits can be solved by using trigonometric identities to rewrite the expression and then evaluating the limit.
It is important to consider these special cases when solving trigonometric limits without using the L'Hospital Rule.

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