Exercise Evaluate the following limit:
lim x → ∞ 8 x ( 4 + 2 x ) 3 x 2 \lim _ { x \rightarrow \infty} \frac{8^x}{{(4+\frac{2}{x})}^{\frac{3x}{2}}} x → ∞ lim ( 4 + x 2 ) 2 3 x 8 x
Final Answer
Show final answer
lim x → ∞ 8 x ( 4 + 2 x ) 3 x 2 = e − 3 4 \lim _ { x \rightarrow \infty} \frac{8^x}{{(4+\frac{2}{x})}^{\frac{3x}{2}}}=e^{-\frac{3}{4}} x → ∞ lim ( 4 + x 2 ) 2 3 x 8 x = e − 4 3
Solution
First, we try to plug in x = ∞ x = \infty x = ∞
8 ∞ ( 4 + 2 ∞ ) 3 ∞ 2 \frac{8^\infty}{{(4+\frac{2}{\infty})}^{\frac{3\infty}{2}}} ( 4 + ∞ 2 ) 2 3 ∞ 8 ∞
We got the phrase ∞ ∞ \frac{\infty}{\infty} ∞ ∞ indeterminate form
We will simplify the phrase using Powers and Roots Rules
lim x → ∞ 8 x ( 4 + 2 x ) 3 x 2 = \lim _ { x \rightarrow \infty} \frac{8^x}{{(4+\frac{2}{x})}^{\frac{3x}{2}}}= x → ∞ lim ( 4 + x 2 ) 2 3 x 8 x =
= lim x → ∞ 2 3 x ( 4 ( 1 + 2 4 x ) ) 3 x 2 = =\lim _ { x \rightarrow \infty} \frac{2^{3x}}{{(4(1+\frac{2}{4x}))}^{\frac{3x}{2}}}= = x → ∞ lim ( 4 ( 1 + 4 x 2 ) ) 2 3 x 2 3 x =
= lim x → ∞ 2 3 x ( 2 2 ) 3 x 2 ( 1 + 1 2 x ) 3 x 2 = =\lim _ { x \rightarrow \infty} \frac{2^{3x}}{{(2^2)}^{\frac{3x}{2}}{(1+\frac{1}{2x})}^{\frac{3x}{2}}}= = x → ∞ lim ( 2 2 ) 2 3 x ( 1 + 2 x 1 ) 2 3 x 2 3 x =
= lim x → ∞ 2 3 x 2 3 x ( 1 + 1 2 x ) 3 x 2 = =\lim _ { x \rightarrow \infty} \frac{2^{3x}}{2^{3x}{(1+\frac{1}{2x})}^{\frac{3x}{2}}}= = x → ∞ lim 2 3 x ( 1 + 2 x 1 ) 2 3 x 2 3 x =
= lim x → ∞ 1 ( 1 + 1 2 x ) 3 x 2 = =\lim _ { x \rightarrow \infty} \frac{1}{{(1+\frac{1}{2x})}^{\frac{3x}{2}}}= = x → ∞ lim ( 1 + 2 x 1 ) 2 3 x 1 =
= lim x → ∞ ( 1 + 1 2 x ) − 3 x 2 = =\lim _ { x \rightarrow \infty}{(1+\frac{1}{2x})}^{-\frac{3x}{2}}= = x → ∞ lim ( 1 + 2 x 1 ) − 2 3 x =
We will plug in infinity again and get
= ( 1 + 1 2 ∞ ) − 3 ∞ 2 = 1 − ∞ ={(1+\frac{1}{2\infty})}^{-\frac{3\infty}{2}}=1^{-\infty} = ( 1 + 2 ∞ 1 ) − 2 3 ∞ = 1 − ∞
We got the phrase 1 ∞ 1^{\infty} 1 ∞ indeterminate form
We will use the known limit also called Euler’s Limit
1 + 1 2 x 1+\frac{1}{2x} 1 + 2 x 1
And the following holds:
lim x → ∞ 1 2 x = 0 \lim _ { x \rightarrow \infty} \frac{1}{2x}= 0 x → ∞ lim 2 x 1 = 0
As required.
Therefore, we will multiply the power by the inverted expression and get
= lim x → ∞ ( 1 + 1 2 x ) 2 x ⋅ 1 2 x ⋅ ( − 3 x 2 ) = =\lim _ { x \rightarrow \infty}{(1+\frac{1}{2x})}^{2x\cdot\frac{1}{2x}\cdot (-\frac{3x}{2})}= = x → ∞ lim ( 1 + 2 x 1 ) 2 x ⋅ 2 x 1 ⋅ ( − 2 3 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 + 1 2 x ) 2 x = e =\lim _ { x \rightarrow \infty}{(1+\frac{1}{2x})}^{2x}=e = x → ∞ lim ( 1 + 2 x 1 ) 2 x = e
All we have left to do is to calculate the limit on the expression that remains in the power:
lim x → ∞ 1 2 x ⋅ ( − 3 x 2 ) = \lim _ { x \rightarrow \infty}\frac{1}{2x}\cdot (-\frac{3x}{2})= x → ∞ lim 2 x 1 ⋅ ( − 2 3 x ) =
= lim x → ∞ − 3 x 4 x = =\lim _ { x \rightarrow \infty}\frac{-3x}{4x}= = x → ∞ lim 4 x − 3 x =
= − 3 4 =-\frac{3}{4} = − 4 3
Therefore, in total we get
= lim x → ∞ ( 1 + 1 2 x ) 2 x ⋅ 1 2 x ⋅ ( − 3 x 2 ) = =\lim _ { x \rightarrow \infty}{(1+\frac{1}{2x})}^{2x\cdot\frac{1}{2x}\cdot (-\frac{3x}{2})}= = x → ∞ lim ( 1 + 2 x 1 ) 2 x ⋅ 2 x 1 ⋅ ( − 2 3 x ) =
= e − 3 4 =e^{-\frac{3}{4}} = e − 4 3
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