$$ \text { Let the function } f(x)=\left\{\begin{array}{cl} \frac{\log _{e}(1+5 x)-\log _{e}(1+\alpha x)}{x} & ;\text { if } x \neq 0 \\ 10 & ; \text { if } x=0 \end{array} \text { be continuous at } x=0 .\right. $$
Then $\alpha$ is equal to
Solution
<p>$f(x)$ is continuous at $x = 0$</p>
<p>$\therefore$ $f(0) = \mathop {\lim }\limits_{x \to 0} f(x)$</p>
<p>$$ \Rightarrow 10 = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}(1 + 5x) - {{\log }_e}(1 + \alpha x)} \over x}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {{\log (1 + 5x)} \over {5x}} \times 5 - {{{{\log }_e}(1 + \alpha x)} \over {\alpha x}} \times \alpha $$</p>
<p>$= 1 \times 5 - \alpha$</p>
<p>$\Rightarrow \alpha = 5 - 10 = - 5$</p>
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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