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Exercise 9.1 · Q3
Q.

Complete the table using a calculator and use the result to estimate the limit.

lim⁡x→0x+3−3x\lim_{x\to0}\dfrac{\sqrt{x+3}-\sqrt3}{x}

xx−0.1-0.1−0.01-0.01−0.001-0.0010.0010.0010.010.010.10.1
f(x)f(x)
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✓ Free question

Step 1. Try direct substitution. At x=0x=0: 3−30=00\dfrac{\sqrt3-\sqrt3}{0}=\dfrac00, indeterminate.

Step 2. Rationalize the numerator. Multiply numerator and denominator by the conjugate x+3+3\sqrt{x+3}+\sqrt3:

x+3−3x⋅x+3+3x+3+3=(x+3)−3x(x+3+3)=xx(x+3+3).\dfrac{\sqrt{x+3}-\sqrt3}{x}\cdot\dfrac{\sqrt{x+3}+\sqrt3}{\sqrt{x+3}+\sqrt3}=\dfrac{(x+3)-3}{x\left(\sqrt{x+3}+\sqrt3\right)}=\dfrac{x}{x\left(\sqrt{x+3}+\sqrt3\right)}.

Step 3. Cancel xx (valid for x≠0x\ne0).

=1x+3+3.=\dfrac1{\sqrt{x+3}+\sqrt3}.

Step 4. Table (using the simplified form).

xx−0.1-0.1−0.01-0.01−0.001-0.0010.0010.0010.010.010.10.1
f(x)f(x)0.28910.28910.28860.28860.28870.28870.28870.28870.28870.28870.28840.2884

Both sides trend to about 0.28870.2887.

Step 5. Take the limit. lim⁡x→01x+3+3=13+3=123=36≈0.2887\displaystyle\lim_{x\to0}\dfrac1{\sqrt{x+3}+\sqrt3}=\dfrac1{\sqrt3+\sqrt3}=\dfrac1{2\sqrt3}=\dfrac{\sqrt3}{6}\approx0.2887, matching the table.

✓Final answer

lim⁡x→0x+3−3x=36\displaystyle\lim_{x\to0}\dfrac{\sqrt{x+3}-\sqrt3}{x}=\boxed{\dfrac{\sqrt3}{6}}

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