Resolved: "How to evaluate the limit of this expression?limx→∞sin2(x+1−x)1−cos21x"

Maximillian Patterson

Answered question

2022-04-20

How to evaluate the limit of this expression?

lim_{x \to \infty} \frac{\sin^{2} (\sqrt{x + 1} - \sqrt{x})}{1 - \cos^{2} \frac{1}{x}}

Alexis Wolf

Beginner2022-04-21Added 13 answers

Observe

lim_{x \to \infty} \frac{\sin^{2} (\sqrt{x + 1} - \sqrt{x})}{1 - \cos^{2} \frac{1}{x}} = lim_{x \to \infty} \frac{\sin^{2} (\sqrt{x + 1} - \sqrt{x})}{\sin^{2} \frac{1}{x}}

= lim_{x \to \infty} \frac{{(\sqrt{x + 1} - \sqrt{x})}^{2} \frac{\sin^{2} (\sqrt{x + 1} - \sqrt{x})}{{(\sqrt{x + 1} - \sqrt{x})}^{2}}}{\frac{1}{x^{2}} \frac{\sin^{2} (\frac{1}{x})}{\frac{1}{x^{2}}}}

Now, setting

t = \frac{1}{\sqrt{x + 1} + \sqrt{x}}

we note

x \to \infty

implies

t \to 0^{+}

, and since

lim_{t \to 0} \frac{\sin t}{t} = 1

lim_{x \to \infty} \frac{\sin^{2} (\sqrt{x + 1} - \sqrt{x})}{{(\sqrt{x + 1} - \sqrt{x})}^{2}} = {[lim_{x \to \infty} \frac{\sin \frac{1}{\sqrt{x + 1} + \sqrt{x}}}{\frac{1}{\sqrt{x + 1} + \sqrt{x}}}]}^{2}

= {(lim_{t \to 0^{+}} \frac{\sin t}{t})}^{2} = {(1)}^{2} = 1

Similarly, putting

t = \frac{1}{x}

we have

t \to 0^{+}

x \to \infty

, and

lim_{x \to \infty} \frac{\sin^{2} (\frac{1}{x})}{\frac{1}{x^{2}}} = {(lim_{t \to 0^{+}} \frac{\sin t}{t})}^{2} = 1^{2} = 1

Then,

lim_{x \to \infty} \frac{\sin^{2} (\sqrt{x + 1} - \sqrt{x})}{1 - \cos^{2} \frac{1}{x}} = (lim_{x \to \infty} \frac{{(\sqrt{x + 1} - \sqrt{x})}^{2}}{\frac{1}{x^{2}}}) lim_{x \to \infty} \frac{\frac{\sin^{2} (\sqrt{x + 1} - \sqrt{x})}{{(\sqrt{x + 1} - \sqrt{x})}^{2}}}{\frac{\sin^{2} (\frac{1}{x})}{\frac{1}{x^{2}}}}

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

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

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