Limits without L'Hospital's rule

Hey guys,

today during class my lecturer mentioned that some limits that L'Hospital's rule applies to and seem to be solvable only by using the rule, can be in fact used without differentiation. Could you give any hints on the example he gave us?

$\displaystyle $\lim _{x\to 0 }{ 5x^2 - tg x\over x} $$

Re: Limits without L'Hospital's rule

Most limits can (and should) be found by expanding the function into its Taylor series. For example, here sin(x) = x + O(x^3) using the big-O notation.

Re: Limits without L'Hospital's rule

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Originally Posted by

**Doubled144314** Hey guys,

today during class my lecturer mentioned that some limits that L'Hospital's rule applies to and seem to be solvable only by using the rule, can be in fact used without differentiation. Could you give any hints on the example he gave us?

$\displaystyle $\lim _{x\to 0 }{ 5x^2 - tg x\over x} $$

$\displaystyle \lim _{x\to 0 } \frac{ 5x^2 - \tan{x}}{x} =$

$\displaystyle \lim _{x\to 0 } 5x - \frac{\sin{x}}{x} \cdot \frac{1}{\cos{x}}$

Re: Limits without L'Hospital's rule

Quote:

Originally Posted by

**Doubled144314** Hey guys,

today during class my lecturer mentioned that some limits that L'Hospital's rule applies to and seem to be solvable only by using the rule, can be in fact used without differentiation. Could you give any hints on the example he gave us?

$\displaystyle $\lim _{x\to 0 }{ 5x^2 - tg x\over x} $$

Since x technically isn't 0, you could just divide out the x and get 5x - tg.

Re: Limits without L'Hospital's rule

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Originally Posted by

**phys251** Since x technically isn't 0, you could just divide out the x and get 5x - tg.

Definitely not! tg(x) is another way to write tan(x). tan(x)/x is NOT equal to tan...

I personally see no advantage to evaluating limits using Taylor expansions as opposed to L'Hospital's Rule, as these require differentiation to be found also. You are expected to be able to perform some algebraic manipulation so that you can get the function into a form where the limits can be evaluated, as in Skeeter's post.

Re: Limits without L'Hospital's rule

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Originally Posted by

**Prove It** I personally see no advantage to evaluating limits using Taylor expansions as opposed to L'Hospital's Rule, as these require differentiation to be found also.

For most functions occurring in the expression whose limit has to be found, Taylor expansions are known, so no differentiation is required. One advantage of this method is that you know the speed of convergence. For example, (sin(x) - x) / x tends to 0 as O(x²).

Re: Limits without L'Hospital's rule

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Originally Posted by

**emakarov** For most functions occurring in the expression whose limit has to be found, Taylor expansions are known, so no differentiation is required. One advantage of this method is that you know the speed of convergence. For example, (sin(x) - x) / x tends to 0 as O(x²).

Taylor Expansions are known ONLY to those who are able to evaluate derivatives.

Re: Limits without L'Hospital's rule

Thanks for clearing that up for me guys. I have another question: could you recommend any books to study Calculus in depth? I presume older textbooks are more precise and to the point. My calculus textbook barely mentions Taylor's series and does not have any challenging problems.