For all even integral value of n, lim (x+1)^-n
n to ∞
Exist or not
True or false with full explanation
limn→∞(x+1)−n\mathop {\lim }\limits_{n \to \infty } {\left( {x + 1} \right)^{ - n}}n→∞lim(x+1)−n
=limn→∞exp(ln((x+1)−n))= \mathop {\lim }\limits_{n \to \infty } \exp \left( {\ln \left( {{{\left( {x + 1} \right)}^{ - n}}} \right)} \right)=n→∞limexp(ln((x+1)−n))
=limn→∞exp(−nln(x+1))= \mathop {\lim }\limits_{n \to \infty } \exp \left( { - n\ln \left( {x + 1} \right)} \right)=n→∞limexp(−nln(x+1))
=exp(−ln(x+1)⋅limn→∞n)= \exp \left( { - \ln \left( {x + 1} \right) \cdot \mathop {\lim }\limits_{n \to \infty } n} \right)=exp(−ln(x+1)⋅n→∞limn)
Now for ln(x+1)>0\ln \left( {x + 1} \right) > 0ln(x+1)>0
=exp(−ln(x+1)⋅∞)= \exp \left( { - \ln \left( {x + 1} \right) \cdot \infty } \right)=exp(−ln(x+1)⋅∞)
=exp(−∞)= \exp \left( { - \infty } \right)=exp(−∞)
=0=0=0
Hence limn→∞(x+1)−n=0\mathop {\lim }\limits_{n \to \infty } {\left( {x + 1} \right)^{ - n}}=0n→∞lim(x+1)−n=0 for ln(x+1)>0\ln \left( {x + 1} \right) > 0ln(x+1)>0
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