Question

If $\displaystyle \lim_{x \to \frac{1}{2}}$$\frac{32x^{5}-1}{2x-1}$ $=\displaystyle \lim_{\theta \to 0}$$\frac{tan\left(\theta / m\right)}{\theta}$, then $m$ is equal to

• A. $2$
• B. $5$
• C. $\frac{1}{5}$
• D. $\frac{1}{2}$

Answer 1

The Correct Option is C

$\displaystyle \lim_{x \to \frac{1}{2}}$$\frac{32x^{5}-1}{2x-1}$ $=\displaystyle \lim_{\theta \to 0}$$\frac{tan\left(\theta / m\right)}{\theta}$ $\Rightarrow \left(\frac{32}{2}\right)$ $\displaystyle \lim_{x \to \frac{1}{2}}$$\frac{x^{5}-\frac{1}{32}}{x-\frac{1}{2}}=\frac{1}{m}$ $\displaystyle \lim_{\theta \to 0}$$\left(\frac{tan \frac{\theta}{m}}{\frac{\theta}{m}}\right)$ $\Rightarrow 16$ $\displaystyle \lim_{x \to \frac{1}{2}}$$\frac{x^{5}-\left(\frac{1}{2}\right)^{5}}{x-\frac{1}{2}}=\frac{1}{m}\times 1$ $\Rightarrow 16 \times5 \times\left(\frac{1}{2}\right)^{5-1}=\frac{1}{m}$ $\Rightarrow 16\times5\times\frac{1}{16}=\frac{1}{m}$ or $m=\frac{1}{5}$