"Find the limit of (e2x+1)1x" was answered by our experts

Aileen Rogers

Answered question

2022-04-12

Find the limit of

{(e^{2 x} + 1)}^{\frac{1}{x}}

shanna87mn

Beginner2022-04-13Added 19 answers

We have

e^{2 x} < e^{2 x} + 1 < e^{2 x} + e^{2 x}

for

x > 0

. Hence, we get

{(e^{2 x})}^{\frac{1}{x}} < {(e^{2 x} + 1)}^{\frac{1}{x}} < {(e^{2 x} + e^{2 x})}^{\frac{1}{x}}

i.e.

e^{2} < {(e^{2 x} + 1)}^{\frac{1}{x}} < {(2 e^{2 x})}^{\frac{1}{x}} = 2^{\frac{1}{x}} e^{2}

Now take the limit and note that

lim_{x \to \infty} 2^{\frac{1}{x}} = 1

to get that

e^{2} \leq lim_{x \to \infty} {(e^{2 x} + 1)}^{\frac{1}{x}} \leq e^{2}

Hence, we get that

lim_{x \to \infty} {(e^{2 x} + 1)}^{\frac{1}{x}} = e^{2}

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