Evaluate: _ x 0 (1+x)^ 6 -1 (1+x)^ 2 -1

Question

Evaluate:
$\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{6}-1}{(1+\text{x})^{2}-1}$

✓

Answer

Given that $\lim\limits_{\text{x} \rightarrow 0}\frac{(1+\text{x})^{6}-1}{(1+\text{x})^{2}-1}$
Dividing the numerator and denominator by x, we get
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{(1+\text{x})^{6}-1}{\text{x}}}{\frac{(1+\text{x})^{2}-1}{\text{x}}}$
Putting 1 + x = y
⇒ x = y - 1
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{\text{y}^{6}-(1)^{6}}{\text{y}-1}}{\frac{\text{y}^{2}-(1)^{2}}{\text{y}-1}}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{\frac{\text{y}^{6}-(1)^{6}}{\text{y}-1}}{\frac{\text{y}^{2}-(1)^{2}}{\text{y}-1}}$
$\frac{6.(1)^{6-1}}{2.(1)^{2-1}}=\frac{6}{2}$
$=3$
Hence, the required qnswer is 3.

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