Question 135

$$\lim_{x \rightarrow 1+} \frac{x^2 - \sqrt{x}}{\sqrt{x} - 1} =$$

Solution

$$=\frac{x^2 - \sqrt{x}}{\sqrt{x} - 1}$$

After differentiation:

$$\frac{2x - \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}$$

Now,

= $$\lim_{x \rightarrow 1}\frac{2x - \frac{1}{2\sqrt{x}}}{\frac{1}{2\sqrt{x}}}$$

= $$\frac{2 - \frac{1}{2}}{\frac{1}{2}}$$

= $$\frac{\frac{3}{2}}{\frac{1}{2}}$$

= $$3$$

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