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f(x)=(2sin(x))/x−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

lim⁡x→0(2sin⁡(x)x)\lim_{x\to0}\left(\frac{2\sin\left(x\right)}{x}\right)

Step-by-step Solution

1

Applying the Sandwich Theorem, which states that: Let II be an interval that contains the point cc, and let f(x)f(x), g(x)g(x), and h(x)h(x) be functions defined on II. If for every xx not equal to cc in the interval II we have g(x)≤f(x)≤h(x)g(x)\leq f(x)\leq h(x) and also suppose that: lim⁡x→cg(x)=lim⁡x→ch(x)=L\displaystyle\lim_{x\to c}{g(x)}=\lim_{x\to c}{h(x)}=L, then: lim⁡x→cf(x)=L\displaystyle\lim_{x\to c}{f(x)}=L

22

Final answer to the exercise

22

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Main Topic: Limits by Direct Substitution

Find limits of functions at a specific point by directly plugging the value into the function.

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limx→0(2sin(x)x )
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