What is the limit as x approaches 0 of 7csc(x)-7/x ?

The limit as x approaches 0 of 7csc(x)-7/x is 0

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limit as x approaches 0 of 7csc(x)-7/x

x \to 0 lim (7 csc (x) - \frac{7}{x})

0

Solution steps

x \to 0 lim (7 csc (x) - \frac{7}{x})

7 csc (x) - \frac{7}{x} = 7 csc (x) (1 - \frac{\frac{7}{x}}{7 csc ( x )})

= x \to 0 lim (7 csc (x) (1 - \frac{\frac{7}{x}}{7 csc ( x )}))

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 7 \cdot x \to 0 lim (csc (x) (1 - \frac{\frac{7}{x}}{7 csc ( x )}))

Simplify

csc (x) (1 - \frac{\frac{7}{x}}{7 csc ( x )}) : csc (x) - \frac{1}{x}

= 7 \cdot x \to 0 lim (csc (x) - \frac{1}{x})

Express with sin, cos

= 7 \cdot x \to 0 lim (\frac{1}{sin ( x )} - \frac{1}{x})

Simplify

\frac{1}{sin ( x )} - \frac{1}{x} : \frac{x - sin ( x )}{x sin ( x )}

= 7 \cdot x \to 0 lim (\frac{x - sin ( x )}{x sin ( x )})

Apply L'Hopital's Rule

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

= 7 \cdot x \to 0 lim (\frac{1 - cos ( x )}{sin ( x ) + x cos ( x )})

Apply L'Hopital's Rule

: x \to 0 lim (\frac{sin ( x )}{2 cos ( x ) - x sin ( x )})

= 7 \cdot x \to 0 lim (\frac{sin ( x )}{2 cos ( x ) - x sin ( x )})

Plug in the value

x = 0

= 7 \cdot \frac{sin ( 0 )}{2 cos ( 0 ) - 0 \cdot sin ( 0 )}

Simplify

7 \cdot \frac{sin ( 0 )}{2 cos ( 0 ) - 0 \cdot sin ( 0 )} : 0

= 0

y^{^{''}} - (\frac{4}{x}) y^{^{'}} - (\frac{8}{x ^{2}}) y = 0

\int x (4 - 2 x^{2}) d x

y = \frac{6}{x ^{4}} - \frac{4}{x ^{2}} - \frac{3}{x}

\frac{\partial}{\partial x} (x^{2} e^{\frac{y}{x}})

\int 9 x^{2} - 4 x + 1 d x

What is the limit as x approaches 0 of 7csc(x)-7/x ?
The limit as x approaches 0 of 7csc(x)-7/x is 0