Processing math: 100%
 
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avatar+1832 

 Nov 27, 2014

Best Answer 

 #2
avatar+33657 
+15

The first one is, perhaps, more complicated than necessary:

 

limx0(cotx1x)

 

limx0(cosxsinx1x)

 

As x → 0, cos x → 1 and sin x → x, so we get:

 

limx0(1x1x)=limx0(0)=0

.

The second one is ok.

.

 Nov 27, 2014
 #1
avatar+1832 
0

Also this 

 

 Nov 27, 2014
 #2
avatar+33657 
+15
Best Answer

The first one is, perhaps, more complicated than necessary:

 

limx0(cotx1x)

 

limx0(cosxsinx1x)

 

As x → 0, cos x → 1 and sin x → x, so we get:

 

limx0(1x1x)=limx0(0)=0

.

The second one is ok.

.

Alan Nov 27, 2014
 #3
avatar+1832 
0

thank you alan 

 Nov 28, 2014
 #4
avatar+1832 
0

What about this how can I solve it 

 

 Nov 28, 2014
 #5
avatar+1832 
0

And for the second picture I found that when x goes to infinity fx goes to infinity ! 

but the graph doesn't show that !!! 

 

https://www.desmos.com/calculator/f49yopuqer

 Nov 28, 2014
 #6
avatar+33657 
+5

limx0(x+1x2sin(2x))

 

Write this as:

limx0(1+1x2sin(2x))

 

as x tends to 0, sin(2x) tends to 2x, so 2/2x tends to 1/x:

limx0(1+1x1x)=1

.

The graph of ex/x4 does go to infinity as x goes to infinity.  It does so so rapidly, that the values get too large to show for the range of x's you can see!

.

 Nov 28, 2014
 #7
avatar+1832 
0

thank you 

 Nov 28, 2014

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