How do you find the limit of $\frac{sin x}{2 x^{2} - x}$ $sinx/(2x^2-x)$ as x approaches 0?

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Answer 1 · 2016-01-25T19:49:29

$- 1$ $-1$

One thing you should know going into this is that

$lim_(xrarr0)sinx/x=1$

This is an important identity in limit problems.

Keeping this in mind, we can factor an $x$ from the denominator of the fraction, giving

$=lim_(xrarr0)(sinx)/(x(2x-1)$

We can rearrange this to get $sinx/x$ , which we already know the limit of.

$=lim_(xrarr0)(sinx/x)1/(2x-1)$

Limits can be multiplied, as follows:

$=lim_(xrarr0)sinx/x*lim_(xrarr0)1/(2x-1)$

Since $lim_(xrarr0)sinx/x=1$ , the expression simply equals

$=lim_(xrarr0)1/(2x-1)$

And here, we can evaluate the limit straightaway by plugging in $0$ for $x$ .

$=1/(2(0)-1)=-1$

We can check the graph at $x=0$ :

graph{sinx/(2x^2-x) [-5.206, 5.89, -2.9, 2.65]}

How do you find the limit of sinx/(2x^2-x)sinx2x2−xsinx/(2x^2-x) as x approaches 0?