Exercise Evaluate the following limit:
lim x → ∞ ln ( 1 + 2 x ) ln ( 1 + 3 x ) \lim _ { x \rightarrow \infty} \ln (1+2^x)\ln (1+\frac{3}{x}) x → ∞ lim ln ( 1 + 2 x ) ln ( 1 + x 3 )
Final Answer
Show final answer
lim x → ∞ ln ( 1 + 2 x ) ln ( 1 + 3 x ) = 3 ln 2 \lim _ { x \rightarrow \infty} \ln (1+2^x)\ln (1+\frac{3}{x})=3\ln 2 x → ∞ lim ln ( 1 + 2 x ) ln ( 1 + x 3 ) = 3 ln 2
Solution
First, we try to plug in x = ∞ x = \infty x = ∞
ln ( 1 + 2 ∞ ) ln ( 1 + 3 ∞ ) = ∞ ⋅ 0 \ln (1+2^\infty)\ln (1+\frac{3}{\infty})=\infty\cdot 0 ln ( 1 + 2 ∞ ) ln ( 1 + ∞ 3 ) = ∞ ⋅ 0
We got the phrase 0 ⋅ ∞ 0\cdot\infty 0 ⋅ ∞ indeterminate form
Using Logarithm Rules we get
lim x → ∞ ln ( 1 + 2 x ) ln ( 1 + 3 x ) = \lim _ { x \rightarrow \infty} \ln (1+2^x)\ln (1+\frac{3}{x})= x → ∞ lim ln ( 1 + 2 x ) ln ( 1 + x 3 ) =
= lim x → ∞ ln ( 1 + 3 x ) ln ( 1 + 2 x ) = =\lim _ { x \rightarrow \infty} \ln {(1+\frac{3}{x})}^{\ln (1+2^x)}= = x → ∞ lim ln ( 1 + x 3 ) l n ( 1 + 2 x ) =
= ln lim x → ∞ ( 1 + 3 x ) ln ( 1 + 2 x ) = =\ln \lim _ { x \rightarrow \infty} {(1+\frac{3}{x})}^{\ln (1+2^x)}= = ln x → ∞ lim ( 1 + x 3 ) l n ( 1 + 2 x ) =
Note: One can insert the boundary because ln function is a continuous function.
We will use the known limit also called Euler’s Limit
1 + 3 x 1+\frac{3}{x} 1 + x 3
And the following holds
lim x → ∞ 3 x = 0 \lim _ { x \rightarrow \infty}\frac{3}{x}= 0 x → ∞ lim x 3 = 0
As required.
Therefore, we will multiply the power by the inverted expression and get
= ln lim x → ∞ ( 1 + 3 x ) x 3 ⋅ 3 x ⋅ ln ( 1 + 2 x ) = =\ln \lim _ { x \rightarrow \infty} {(1+\frac{3}{x})}^{\frac{x}{3}\cdot\frac{3}{x}\cdot\ln (1+2^x)}= = ln x → ∞ lim ( 1 + x 3 ) 3 x ⋅ x 3 ⋅ l n ( 1 + 2 x ) =
Note: When adding an expression by multiplication, its inverted expression must also be multiplied so that the original expression does not change.
Now, according to Euler’s Limit
lim x → ∞ ( 1 + 3 x ) x 3 = e \lim _ { x \rightarrow \infty} {(1+\frac{3}{x})}^{\frac{x}{3}}=e x → ∞ lim ( 1 + x 3 ) 3 x = e
All we have left to do is to calculate the limit on the expression that remains in the power:
lim x → ∞ 3 x ⋅ ln ( 1 + 2 x ) = \lim _ { x \rightarrow \infty}\frac{3}{x}\cdot\ln (1+2^x)= x → ∞ lim x 3 ⋅ ln ( 1 + 2 x ) =
= lim x → ∞ 3 ln ( 1 + 2 x ) x = =\lim _ { x \rightarrow \infty}\frac{3\ln (1+2^x)}{x}= = x → ∞ lim x 3 ln ( 1 + 2 x ) =
We plug in infinity and get
= 3 ln ( 1 + 2 ∞ ) ∞ = ∞ ∞ =\frac{3\ln (1+2^\infty)}{\infty}=\frac{\infty}{\infty} = ∞ 3 ln ( 1 + 2 ∞ ) = ∞ ∞
We got the phrase ∞ ∞ \frac{\infty}{\infty} ∞ ∞ indeterminate form Lopital Rule
= 3 lim x → ∞ 1 1 + 2 x ⋅ 2 x ⋅ ln 2 1 = =3\lim _ { x \rightarrow \infty}\frac{\frac{1}{1+2^x}\cdot 2^x\cdot\ln 2}{1}= = 3 x → ∞ lim 1 1 + 2 x 1 ⋅ 2 x ⋅ ln 2 =
= 3 ln 2 lim x → ∞ 2 x 1 + 2 x = =3\ln 2\lim _ { x \rightarrow \infty}\frac{2^x}{1+2^x}= = 3 ln 2 x → ∞ lim 1 + 2 x 2 x =
Plugging in infinity gives the phrase ∞ ∞ \frac{\infty}{\infty} ∞ ∞ 2 x 2^x 2 x
= 3 ln 2 lim x → ∞ 1 1 2 x + 1 = =3\ln 2\lim _ { x \rightarrow \infty}\frac{1}{\frac{1}{2^x}+1}= = 3 ln 2 x → ∞ lim 2 x 1 + 1 1 =
We plug in infinity again and this time we get
= 3 ln 2 ⋅ 1 1 2 ∞ + 1 = =3\ln 2\cdot\frac{1}{\frac{1}{2^\infty}+1}= = 3 ln 2 ⋅ 2 ∞ 1 + 1 1 =
= 3 ln 2 ⋅ 1 1 ∞ + 1 = =3\ln 2\cdot\frac{1}{\frac{1}{\infty}+1}= = 3 ln 2 ⋅ ∞ 1 + 1 1 =
= 3 ln 2 ⋅ 1 0 + 1 = =3\ln 2\cdot\frac{1}{0+1}= = 3 ln 2 ⋅ 0 + 1 1 =
= 3 ln 2 =3\ln 2 = 3 ln 2
Therefore, in total we get
= ln lim x → ∞ ( 1 + 3 x ) x 3 ⋅ 3 x ⋅ ln ( 1 + 2 x ) = =\ln \lim _ { x \rightarrow \infty} {(1+\frac{3}{x})}^{\frac{x}{3}\cdot\frac{3}{x}\cdot\ln (1+2^x)}= = ln x → ∞ lim ( 1 + x 3 ) 3 x ⋅ x 3 ⋅ l n ( 1 + 2 x ) =
= ln e 3 ln 2 = =\ln e^{3\ln 2}= = ln e 3 l n 2 =
= 3 ln 2 =3\ln 2 = 3 ln 2
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