+1 vote
169 views
$\lim_{x\rightarrow \infty }\left(4^{x}+5^{x}\right)^{1/x}$
in Calculus
retagged | 169 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 $\infty^0$.

$\lim_{x \to \infty} (4^x+5^x)^\frac{1}{x} = \lim_{x \to \infty} \left(5^x \left((\frac{4}{5})^x +1\right )\right)^\frac{1}{x}\\ =\lim_{x \to \infty} 5\times( )^{\frac{1}{\infty}} \text {(the term inside () ranges between 1 and 2)} \\ =5\times()^0\\ =5\times1\\=5$
by Junior (985 points)
selected by
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. $0^0 = 1$.

+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
by (269 points)
$\bf{\displaystyle \lim_{x\rightarrow \infty} \left(4^x+5^x \right) \approx \lim_{x\rightarrow \infty}5^x}$

$\bf{So\; \displaystyle \lim_{x\rightarrow \infty} \left(4^x+5^x \right)^{\frac{1}{x}} \approx \lim_{x\rightarrow}\left(5\right)^{\frac{x}{x}} = 5.}$
by Junior (521 points)

+1 vote