+0  
 
0
386
7
avatar+1832 

xvxvxv  Nov 27, 2014

Best Answer 

 #2
avatar+27057 
+15

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

 

$$\lim_{x\rightarrow 0}(\cot x - \frac{1}{x})$$

 

$$\lim_{x \rightarrow 0}(\frac{\cos x}{\sin x}-\frac{1}{x})$$

 

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

 

$$\lim_{x \rightarrow 0}(\frac{1}{x} - \frac{1}{x})=\lim_{x \rightarrow 0}(0) = 0$$

.

The second one is ok.

.

Alan  Nov 27, 2014
 #1
avatar+1832 
0

Also this 

 

xvxvxv  Nov 27, 2014
 #2
avatar+27057 
+15
Best Answer

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

 

$$\lim_{x\rightarrow 0}(\cot x - \frac{1}{x})$$

 

$$\lim_{x \rightarrow 0}(\frac{\cos x}{\sin x}-\frac{1}{x})$$

 

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

 

$$\lim_{x \rightarrow 0}(\frac{1}{x} - \frac{1}{x})=\lim_{x \rightarrow 0}(0) = 0$$

.

The second one is ok.

.

Alan  Nov 27, 2014
 #3
avatar+1832 
0

thank you alan 

xvxvxv  Nov 28, 2014
 #4
avatar+1832 
0

What about this how can I solve it 

 

xvxvxv  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

xvxvxv  Nov 28, 2014
 #6
avatar+27057 
+5

$$\lim_{x \rightarrow 0}(\frac{x+1}{x}-\frac{2}{sin(2x)})$$

 

Write this as:

$$\lim_{x \rightarrow 0}(1+\frac{1}{x}-\frac{2}{sin(2x)})$$

 

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

$$\lim_{x \rightarrow 0}(1+\frac{1}{x}-\frac{1}{x})=1$$

.

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!

.

Alan  Nov 28, 2014
 #7
avatar+1832 
0

thank you 

xvxvxv  Nov 28, 2014

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