# Find the limit. lim_{xrightarrow0^+}frac{sqrt x}{sqrt{sin x}}

Burhan Hopper 2020-11-23 Answered
Find the limit.
$\underset{x\to {0}^{+}}{lim}\frac{\sqrt{x}}{\sqrt{\mathrm{sin}x}}$
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## Expert Answer

Clelioo
Answered 2020-11-24 Author has 88 answers
Given:
$\underset{x\to {0}^{+}}{lim}\frac{\sqrt{x}}{\sqrt{\mathrm{sin}x}}$
Solve:
$\underset{x\to {0}^{+}}{lim}\frac{\sqrt{x}}{\sqrt{\mathrm{sin}x}}=\underset{x\to {0}^{+}}{lim}\sqrt{\frac{\frac{1}{\mathrm{sin}x}}{x}}$
$=\sqrt{\frac{1}{\underset{x\to {0}^{+}}{lim}\frac{\mathrm{sin}x}{x}}}$
$=\sqrt{\frac{1}{\underset{x\to {0}^{+}}{lim}\frac{\mathrm{sin}x}{x}}}$
taking $\underset{x\to {0}^{+}}{lim}\frac{\mathrm{sin}x}{x}$
If substituting in the limit directly, it is reduced to the indeterminate form, so apply LHospitalss rule
$\underset{x\to {0}^{+}}{lim}\frac{\frac{d\mathrm{sin}x}{dx}}{\frac{dx}{dx}}=\underset{x\to {0}^{+}}{lim}\frac{\mathrm{cos}x}{1}=\frac{1}{1}=1$
Equation (1) becomes
$\underset{x\to {0}^{+}}{lim}\sqrt{\frac{x}{\mathrm{sin}x}}=\sqrt{\frac{1}{1}}=1$
The solution is 1.
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