What is the limit as x approaches infinity of (ln(3x+5))/(ln(7x+3)+1) ?

The limit as x approaches infinity of (ln(3x+5))/(ln(7x+3)+1) is 1

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limit as x approaches infinity of (ln(3x+5))/(ln(7x+3)+1)

x \to \infty lim (\frac{ln ( 3 x + 5 )}{ln ( 7 x + 3 ) + 1})

1

Solution steps

x \to \infty lim (\frac{ln ( 3 x + 5 )}{ln ( 7 x + 3 ) + 1})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{\frac{3}{3 x + 5}}{\frac{7}{7 x + 3}})

= x \to \infty lim (\frac{\frac{3}{3 x + 5}}{\frac{7}{7 x + 3}})

Divide fractions:

\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a \cdot d}{b \cdot c}

= x \to \infty lim (\frac{3 ( 7 x + 3 )}{7 ( 3 x + 5 )})

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

= \frac{3}{7} \cdot x \to \infty lim (\frac{7 x + 3}{3 x + 5})

Divide by highest denominator power:

\frac{7 + \frac{3}{x}}{3 + \frac{5}{x}}

= \frac{3}{7} \cdot x \to \infty lim (\frac{7 + \frac{3}{x}}{3 + \frac{5}{x}})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= \frac{3}{7} \cdot \frac{lim _{x \to \infty} ( 7 + \frac{3}{x} )}{lim _{x \to \infty} ( 3 + \frac{5}{x} )}

x \to \infty lim (7 + \frac{3}{x}) = 7

x \to \infty lim (3 + \frac{5}{x}) = 3

= \frac{3}{7} \cdot \frac{7}{3}

Simplify

\frac{3}{7} \cdot \frac{7}{3} : 1

= 1

\frac{d x}{d t} + 7 x = 5 cos (2 t)

n = 1 \sum \infty \frac{n ^{3}}{4 ^{n}}

x \to 1 lim ([g (x) - 3 f (x)])

\int e^{y} d y

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

What is the limit as x approaches infinity of (ln(3x+5))/(ln(7x+3)+1) ?
The limit as x approaches infinity of (ln(3x+5))/(ln(7x+3)+1) is 1