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(D) For the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{e^{\mu x} + 5}{e^{100x} + 7}$ to exist,the degree of the exponential growth in the num

Vedclass · Accepted Answer

$5050$

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(D) For the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{e^{\mu x} + 5}{e^{100x} + 7}$ to exist,the degree of the exponential growth in the numerator must not exceed that of the denominator.Case $1$: If $\mu Case $2$: If $\mu = 100$,then $\mathop {\lim }\limits_{x 	o \infty } \frac{e^{100x} + 5}{e^{100x} + 7} = \mathop {\lim }\limits_{x 	o \infty } \frac{1 + 5e^{-100x}}{1 + 7e^{-100x}} = 1$,which exists.Case $3$: If $\mu > 100$,the limit tends to $\infty$,which does not exist.Thus,$\mu \leq 100$. Since $\mu$ is a positive integer,$\mu \in \{1, 2, 3, \ldots, 100\}$.The sum of these values is $\sum_{k=1}^{100} k = \frac{100(100+1)}{2} = 5050$.

If $\mathop {\lim }\limits_{x \to \infty } \frac{e^{\mu x} + 5}{e^{100x} + 7}$ exists,then the sum of all possible positive integral values of $\mu$ is:

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