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 )

Algebra ->  Test -> 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 )       Log On


   

Question 361546: Please help me solve this equation: Evaluate the following limits or explain why they do not exist.
a. +lim%28x-%3E0%2C+%28xcosx-sinx%29%2Fx+%29+
b. +lim%28x-%3Einfinity%2C+%28e%5Ex-2%5Ex%29%2Fx+%29+
c. +lim%28x-%3E1%2C++%28x%2F%28x-1%29+-+%281%2Flnx%29+%29+%29+
Found 2 solutions by robertb, solver91311:
Answer by robertb(5830) About Me  (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
= lim%28x-%3E0%2C+%28cosx%29%29-lim%28x-%3E0%2C+sinx%2Fx%29 = 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 e%5E0+-+%28ln2%29%2A2%5E0+=+1-ln2.
c. +lim%28x-%3E1%2C++%28x%2F%28x-1%29+-+%281%2Flnx%29+%29+%29+,
=+lim%28x-%3E1%2C+%28+%28xlnx-x%2B1%29%2F%28%28x-1%29%2Alnx%29%29+%29+, combining fractions.
=+lim%28x-%3E1%2C+%28+lnx%2F%28lnx%2B%28x-1%29%2Fx%29%29+%29+, applying L-H rule. (Gives 0/0)
=+lim%28x-%3E1%2C+%28+%28xlnx%29%2F%28xlnx%2Bx-1%29%29%29, simplifying complex fractions, (Gives 0/0)
=+lim%28x-%3E1%2C+%28+%28lnx%2B1%29%2F%28lnx%2B2%29+%29+%29, applying L-H rule (Not indeterminate anymore)
=1%2F2

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!




This is the indeterminate form

so apply L'Hôpital's Rule:




Using the Product and Sum rules on the numerator:



b. Works the same way.

c. You need to combine the two fractions so that you will have a indeterminate form.

John

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