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1 vote
328 views
$\lim_{x\rightarrow \infty }\left(4^{x}+5^{x}\right)^{1/x}$
in Calculus
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328 views
0
if we take 4 common instead of 5 then ans becomes 4 then how we can decide the answer
0
no, answer is 5 only. We can't take 4 out as then the term inside becomes $\infty^0$.

3 Answers

4 votes
 
Best answer
$\lim_{x \to \infty} (4^x+5^x)^\frac{1}{x} = \lim_{x \to \infty} \left(5^x \left((\frac{4}{5})^x +1\right )\right)^\frac{1}{x}\\ =\lim_{x \to \infty} 5\times( )^{\frac{1}{\infty}} \text {(the term inside () ranges between 1 and 2)} \\ =5\times()^0\\ =5\times1\\=5$

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0
() can be anything?
1
Since the power is zero, it will be always equal to 1.
0
So, 0^0 = 1 rt?
1
Sorry, Mistake.
Thanks for pointing out,

$\left(\frac{4}{5} \right )^\infty + 1 \neq 0$

Since $\left( \frac{4}{5}\right)^\infty > 0$
1

Actually you were correct. $0^0 = 1$. 

Ref: http://mathforum.org/library/drmath/view/69917.html

1
Didn't knew that. Thanks for the reference.
0
(4/5)∞ wil always be 0.
1 vote
very easy

lt->inf (4^x+5^x)^1/x

lt->inf 5^x*1/x[(4/5)^x +1]^1/x

ans will be 5

since 4/5 will give <1 no. and its to the power inf will results 0
0 votes
$\bf{\displaystyle \lim_{x\rightarrow \infty} \left(4^x+5^x \right) \approx \lim_{x\rightarrow \infty}5^x}$

$\bf{So\; \displaystyle  \lim_{x\rightarrow \infty} \left(4^x+5^x \right)^{\frac{1}{x}} \approx \lim_{x\rightarrow}\left(5\right)^{\frac{x}{x}} = 5.}$

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