Is f(x) equals to: 2* (sin(x)/x) + (x/2)* sin(1/x). So limit of f(x) when x goes to infinity is equals to:
Is sen the same as sin
Yes ^^
do u know that lim x->0 sin(x)/x=1 ?
Yes, but' In this case, Is x->infinity
well for second term (x/2)* sin(1/x)=0.5sin(1/x)/(1/x) let 1/x=t then t->0
what will u get?
\[ (x/2)* \sin(1/x)= \frac 1 2 \frac {\sin \left(\frac 1 x\right)}{\frac1 x} \]
If x goes to Infinity 1/x goes to zero and the ratio above goes to 1/2
the first term \[ \left |\frac {\sin(x)}{x} \right|\le \frac {1}{|x|} \to 0 \] When x gies to Infinity.
So \[ \lim_{x\to \infty} f(x) =\frac 1 2 \]
Wait, about the second term, it's going to a indeterminate form that equals to (1/2)*(0/0). How this is equals to 1/2?
Aha using L'Hospital discover that this is equals to 1 ^^ thanks!,
@eliassaab I don't understand the first term...Could you, please, explain to me?
Do you know that \( \sin(x)| \le 1 \)
|sin(x)} < = 1
Then \[ \left |\frac {\sin(x)}{x} \right|\le \frac {1}{|x|} \]
If x goes to Infinity then \[ \frac {1}{|x|} \to 0 \]
\[ \left |2\frac {\sin(x)}{x} \right|\le \frac {2}{|x|} \to 0 \]
This about absolute value I didnt knee... So.you tolling me that when |sin(x)/x| goes to zero, sin(x)/x too? Even knowing that the range of the second is -1<=x<=1?
Knew*
Saying in the place of tolling =P
See the following table of {x,f(x)} {{100., 0.489864}, {1100., 0.500779}, {2100., 0.500941}, {3100., 0.500441}, {4100., 0.499893}, {5100., 0.499635}, {6100., 0.499729}, {7100., 0.5}, {8100., 0.500204}, {9100., 0.500204}, {10100., 0.500043}, {11100., 0.499877}, {12100., 0.499837}, {13100., 0.499935}, {14100., 0.500072}, {15100., 0.500132}, {16100., 0.500076}, {17100., 0.499964}, {18100., 0.499894}, {19100., 0.499919}, {20100., 0.500009}, {21100., 0.500083}, {22100., 0.500081}, {23100., 0.500011}, {24100., 0.499938}, {25100., 0.499923}, {26100., 0.499974}, {27100., 0.500043}, {28100., 0.500071}, {29100., 0.500037}}
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