\(\begin{array}{l}\lim_{x \rightarrow 0} \frac{ax-(e^{4x}-1)}{ax(e^{4x}-1)}\end{array} \) Applying L’Hospital Rule \(\begin{array}{l}\lim_{x \rightarrow 0} \frac{a-4e^{4x}}{a(e^{4x}-1)+ax(4e^{4x})}\end{array} \) So for limit to exist,a=4 Applying L’Hospital... View Article
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