SOLUTION: Please help me solve this equation: Evaluate the following limits or explain why they do not exist.
a. {{{ lim(x->0, (xcosx-sinx)/x ) }}}
b. {{{ lim(x->infinity, (e^x-2^x)/x )
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Question 361546: Please help me solve this equation: Evaluate the following limits or explain why they do not exist.
a.
b.
c.
Found 2 solutions by robertb, solver91311:
Answer by robertb(5830) (Show Source): You can put this solution on YOUR website!
a. Putting x = 0 produces the indeterminate form 0/0. But for this problem, there is NO need to use L'Hopital's rule as shown below:
But
= = 1 -1 = 0.
b. Putting x = 0 produces the indeterminate form 0/0.
Now, by direct application of L'Hopital's rule. This does NOT produce an indeterminate form, so the limit is .
c. ,
=, combining fractions.
=, applying L-H rule. (Gives 0/0)
=, simplifying complex fractions, (Gives 0/0)
=, applying L-H rule (Not indeterminate anymore)
=
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
\ -\ \sin(x)}{x})
This is the indeterminate form 
so apply L'Hôpital's Rule:
}{g(x)}\ =\ \lim_{x\rightarrow c}\ \frac{f'(x)}{g'(x)})
Using the Product and Sum rules on the numerator:
\ -\ x\sin(x)\ -\ \cos(x)}{1}\ =\ \lim_{x\rightarrow 0}\ \frac{x\sin(x)}{1}\ =\ 0)
b. Works the same way.
c. You need to combine the two fractions so that you will have a indeterminate form.
John
My calculator said it, I believe it, that settles it
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