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## Homework Statement

(a) Explain why L'Hopital's rule does not apply to the problem

lim[tex]_{x\rightarrow0}[/tex] [ (x[tex]^{2}[/tex]sin(1/x)) / sinx ]

(b) Find the limit.

## Homework Equations

lim [tex]_{x\rightarrow0}[/tex] xsin(1/x) = 0 , by the Squeezing Theorem.

lim [tex]_{x\rightarrow0}[/tex] sin (1/x) Does Not Exist because it oscillates between -1 and 1.

lim [tex]_{x\rightarrow0}[/tex] x[tex]^{2}[/tex]sin(1/x) = 0 by the Squeezing Theorem.

lim[tex]_{x\rightarrow0}[/tex]sinx/x = 1

**3. My attempt(s) at a solution**

I wrote the original problem

lim[tex]_{x\rightarrow0}[/tex] [ (x[tex]^{2}[/tex]sin(1/x)) / sinx ]

as

lim[tex]_{x\rightarrow0}[/tex] sin (1/x) / lim[tex]_{x\rightarrow}[/tex](1/x) * lim[tex]_{x\rightarrow0}[/tex] (sinx/x).

Since the limit of the numerator doesn't exist, and lim[tex]_{x\rightarrow0}[/tex](1/x) is +[tex]\infty[/tex], and lim[tex]_{x\rightarrow0}[/tex] sinx/x = 1, then the limit of the problem doesn't exist, right?