The value of mathop {lim }limits_{n to infty | vedclass

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(C) Given the limit $\mathop {\lim }\limits_{n 	o \infty } \frac{{{x^n}}}{{{x^n} + 1}}$.Since $x  1$,which implies that $\mathop

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(C) Given the limit $\mathop {\lim }\limits_{n 	o \infty } \frac{{{x^n}}}{{{x^n} + 1}}$.Since $x  1$,which implies that $\mathop {\lim }\limits_{n 	o \infty } x^n$ does not exist in the finite sense,but we can divide the numerator and denominator by $x^n$:$\mathop {\lim }\limits_{n 	o \infty } \frac{{{x^n}}}{{{x^n}(1 + \frac{1}{x^n})}} = \mathop {\lim }\limits_{n 	o \infty } \frac{1}{1 + \frac{1}{x^n}}$.As $n 	o \infty$ and $|x| > 1$,$\frac{1}{x^n} 	o 0$.Therefore,the limit is $\frac{1}{1 + 0} = 1$.

The value of $\mathop {\lim }\limits_{n \to \infty } \frac{{{x^n}}}{{{x^n} + 1}}$ where $x < -1$ is

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