What is the limit as x approaches 0 of (x+1)sin(1/x) ?

The limit as x approaches 0 of (x+1)sin(1/x) is diverges

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limit as x approaches 0 of (x+1)sin(1/x)

x \to 0 lim ((x + 1) sin (\frac{1}{x}))

diverges

Solution steps

x \to 0 lim ((x + 1) sin (\frac{1}{x}))

Apply the following algebraic property

: a \cdot b = \frac{a}{\frac{1}{b}}, b \neq = 0

(1 + x) sin (\frac{1}{x}) = \frac{sin ( \frac{1}{x} )}{\frac{1}{( 1 + x )}}

= x \to 0 lim (\frac{sin ( \frac{1}{x} )}{\frac{1}{1 + x}})

Apply Limit Divergence Criterion:

diverges

= diverges

\frac{d}{d x} (cos^{12} (x))

5 t + \frac{7}{t}

\int \frac{1}{3 ^{- x}} d x

\int cos (5 x + 7) d x

f (x) = \frac{x ^{2} - 2}{x ^{2} + 2}

What is the limit as x approaches 0 of (x+1)sin(1/x) ?
The limit as x approaches 0 of (x+1)sin(1/x) is diverges