\lim_{x \\to\infty}(\frac{x+7}{x+2})^{x-5} equal to

NIMCET 2021 Question Paper with solutions

Question : 20 of 120

Marks: +1, -0

\lim ↙{x \→\∞}({x+7}/{x+2})^{x-5}

equal to

Solution:

\lim ↙{x \→\∞}({x+7}/{x+2})^{x-5}

⇒\lim ↙{x \→\∞}[1+{x+7}/{x+2}-1]^{x+5}

⇒e^{ {\lim}↙{x \→\∞}({x+7}/{x+2}-1)^{x+5}}

⇒ e^{{\lim}↙{x \→\∞}{5(x+5)}/{x+2}}

\⇒e^{5}

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