If $$\mathop {\lim }\limits_{x \to 0} {{ax - ({e^{4x}} - 1)} \over {ax({e^{4x}} - 1)}}$$ exists and is equal to b, then the value of a $-$ 2b is __________.
Answer (integer)
5
Solution
$$\mathop {\lim }\limits_{x \to 0} {{ax - \left( {{e^{4x}} - 1} \right)} \over {ax\left( {{e^{4x}} - 1} \right)}}$$<br><br>Applying L' Hospital Rule<br><br>$$\mathop {\lim }\limits_{x \to 0} {{a - 4{e^{4x}}} \over {a\left( {{e^{4x}} - 1} \right) + ax\left( {4{e^{4x}}} \right)}}$$
<br><br>This is ${{a - 4} \over 0}$.
<br><br> limit exist only when $a - 4 = 0$ $\Rightarrow$ a = 4<br><br>Applying L' Hospital Rule<br><br>$$\mathop {\lim }\limits_{x \to 0} {{ - 16{e^{4x}}} \over {a\left( {4{e^{4x}}} \right) + a\left( {4{e^{4x}}} \right) + ax\left( {16{e^{4x}}} \right)}}$$<br><br>= ${{ - 16} \over {4a + 4a}} = {{ - 16} \over {32}} = - {1 \over 2} = b$<br><br>$a - 2b = 4 - 2\left( {{{ - 1} \over 2}} \right) = 4 + 1 = 5$
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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