Burhan Hopper

2020-11-23

Find the limit.

$\underset{x\to {0}^{+}}{lim}\frac{\sqrt{x}}{\sqrt{\mathrm{sin}x}}$

Clelioo

Skilled2020-11-24Added 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.

Solve:

taking

If substituting in the limit directly, it is reduced to the indeterminate form, so apply LHospitalss rule

Equation (1) becomes

The solution is 1.

Jeffrey Jordon

Expert2022-08-24Added 2605 answers

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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