# Limits without L'Hospital's rule

• Dec 18th 2012, 10:38 AM
Doubled144314
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} $$• Dec 18th 2012, 12:02 PM emakarov Re: Limits without L'Hospital's rule Most limits can (and should) be found by expanding the function into its Taylor series. For example, here sin(x) = x + O(x^3) using the big-O notation. • Dec 18th 2012, 01:50 PM skeeter Re: Limits without L'Hospital's rule Quote: Originally Posted by Doubled144314 Hey guys, today during class my lecturer mentioned that some limits that L'Hospital's rule applies to and seem to be solvable only by using the rule, can be in fact used without differentiation. Could you give any hints on the example he gave us? \displaystyle \lim _{x\to 0 }{ 5x^2 - tg x\over x}$$

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

$\displaystyle \lim _{x\to 0 } 5x - \frac{\sin{x}}{x} \cdot \frac{1}{\cos{x}}$
• Dec 18th 2012, 05:05 PM
phys251
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.
• Dec 19th 2012, 12:49 AM
Prove It
Re: Limits without L'Hospital's rule
Quote:

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.
• Dec 19th 2012, 02:08 AM
emakarov
Re: Limits without L'Hospital's rule
Quote:

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).
• Dec 19th 2012, 03:53 AM
Prove It
Re: Limits without L'Hospital's rule
Quote:

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.
• Dec 19th 2012, 06:26 AM
Doubled144314
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.