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Limit of the function log(1+x)/log(2+x)

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     /log(1 + x)\
 lim |----------|
x->oo\log(2 + x)/

\lim_{x \to \infty}\left(\frac{\log{\left(x + 1 \right)}}{\log{\left(x + 2 \right)}}\right)

Limit(log(1 + x)/log(2 + x), x, oo, dir='-')

Lopital's rule

We have indeterminateness of type

oo/oo,

i.e. limit for the numerator is

\lim_{x \to \infty} \log{\left(x + 1 \right)} = \infty

and limit for the denominator is

\lim_{x \to \infty} \log{\left(x + 2 \right)} = \infty

Let's take derivatives of the numerator and denominator until we eliminate indeterninateness.

\lim_{x \to \infty}\left(\frac{\log{\left(x + 1 \right)}}{\log{\left(x + 2 \right)}}\right)

\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \log{\left(x + 1 \right)}}{\frac{d}{d x} \log{\left(x + 2 \right)}}\right)

\lim_{x \to \infty}\left(\frac{x + 2}{x + 1}\right)

\lim_{x \to \infty}\left(\frac{\frac{d}{d x} \left(x + 2\right)}{\frac{d}{d x} \left(x + 1\right)}\right)

\lim_{x \to \infty} 1

\lim_{x \to \infty} 1

1

It can be seen that we have applied Lopital's rule (we have taken derivatives with respect to the numerator and denominator) 2 time(s)

The graph

Plot the graph

Rapid solution [src]

1

Expand and simplify

Other limits x→0, -oo, +oo, 1

\lim_{x \to \infty}\left(\frac{\log{\left(x + 1 \right)}}{\log{\left(x + 2 \right)}}\right) = 1

\lim_{x \to 0^-}\left(\frac{\log{\left(x + 1 \right)}}{\log{\left(x + 2 \right)}}\right) = 0

\lim_{x \to 0^+}\left(\frac{\log{\left(x + 1 \right)}}{\log{\left(x + 2 \right)}}\right) = 0

\lim_{x \to 1^-}\left(\frac{\log{\left(x + 1 \right)}}{\log{\left(x + 2 \right)}}\right) = \frac{\log{\left(2 \right)}}{\log{\left(3 \right)}}

\lim_{x \to 1^+}\left(\frac{\log{\left(x + 1 \right)}}{\log{\left(x + 2 \right)}}\right) = \frac{\log{\left(2 \right)}}{\log{\left(3 \right)}}

\lim_{x \to -\infty}\left(\frac{\log{\left(x + 1 \right)}}{\log{\left(x + 2 \right)}}\right) = 1

The graph