# Show that using the the Squeeze/Sandwich Theorem.

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Show that `lim_(x->0)((x^2+1)sinx)/x=1` :

First note that `lim_(x->0)((x^2+1)sinx)/x=lim_(x->0)(x^2+1)*lim_(x->0)sinx/x`

Now `lim_(x->0)x^2+1=1` by substitution, so we need to show `lim_(x->0)sinx/x=1` .

** Background**

Draw a unit circle. Then draw a right triangle in the first quadrant with the vertex of the right angle at C(1,0), and vertices of the acute angles at A(0,0) and B`(1,tantheta)` where `theta` is the measure of the acute angle at A.

Now `tantheta/2` is the area of that triangle.

(The base is 1, height is `tan theta` )

Consider the area of the sector associated with `theta` : Area is `theta/(2pi)*pir^2` ;since r=1 we have area `theta/2` . This area is smaller than the area of the triangle so `tantheta/2>=theta/2`

Now drop an altitude from the intersection of the hypotenuse and the circle to the x-axis and a segment from that point to (1,0) creating another triangle. The height of this triangle is `sintheta` with a base of 1, so the area of this triangle is `sintheta/2` . This triangle is smaller than the original triangle and the area of the sector.

Now `tantheta/2>=theta/2>=sintheta/2` Multiplying by `2/sintheta` (this is greater than 0) yields:

`1/costheta>=theta/sintheta>=1` Take the reciprocal (reverse the signs) to get

`costheta<=sintheta/theta<=1`

*** `cos(-theta)=costheta` and `sintheta/theta=(sin(-theta))/(-theta)` so the inequality holds for all `theta` .

** End background

`lim_(x->0)cosx=1,lim_(x->0)1=1` so by the squeeze theorem we have `lim_(x->0)(sinx)/x=1`

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`lim_(x->0)((x^2+1)sinx)/x=lim_(x->0)(x^2+1)*lim_(x->0)(sinx)/x=1*1=1` as required.

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The graph of `((x^2+1)sinx)/x` :

**Sources:**

lim x→0 [((x^2)+1)sin(x)]/x = 1 is what I want you guys to show me using the Squeeze/Sandwich Theorem

looks like my copy and paste did not show up :P.