MCQ
અહી વિધેય $f(x)=\left\{\begin{array}{cc}\frac{\log _{e}(1+5 x)-\log _{e}(1+\alpha x)}{x} & \text { if } x \neq 0 \\ 10 & \text {; if } x=0\end{array}\right.$ એ $x=0$ આગળ સતત હોય તો $\alpha$ ની કિમંત મેળવો.
- A$10$
- B$-10$
- C$5$
- D$-5$
$\lim _{x \rightarrow 0} \frac{\ln (1+5 x)-\ln (1+\alpha x)}{x}=10$
Using expension
$\lim _{x \rightarrow 0} \frac{(5 x+\ldots \ldots)-(\alpha x+\ldots \ldots)}{x}=10$
$5-\alpha=10 \Rightarrow \alpha=-5$
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