+0

0
213
7
+1828

xvxvxv  Nov 27, 2014

#2
+26399
+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
Sort:

#1
+1828
0

Also this

xvxvxv  Nov 27, 2014
#2
+26399
+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
#3
+1828
0

thank you alan

xvxvxv  Nov 28, 2014
#4
+1828
0

xvxvxv  Nov 28, 2014
#5
+1828
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
+26399
+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
+1828
0

thank you

xvxvxv  Nov 28, 2014

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