Question

Lim (x->inf) ((x-5)/(x+2))^(2x+10)

Answer 1

$\lim_{x \to \infty} (\frac{x-5}{x+2} )^{2x+10}= \lim_{x \to \infty} (1+\frac{-7}{x+2} )^{\frac{x+2}{-7}*\frac{-7}{x+2}*(2x+10)}=\\ = \lim_{x \to \infty} e^{\frac{-14(x+5)}{x+2}}= e^{ -14\lim_{x \to \infty} \frac{x+5}{x+2}} =e^{-14}$

Answer 2

To find the limit as x approaches infinity of the given expression, let's simplify it step by step.

We have the expression:

lim(x->inf) ((x-5)/(x+2))^(2x+10)

Step 1: Divide numerator and denominator by x As x approaches infinity, both the numerator and denominator grow without bound. To simplify the expression, we can divide both the numerator and denominator by x:

lim(x->inf) ((1 - 5/x) / (1 + 2/x))^(2x+10)

Step 2: Simplify the fraction inside the exponent As x approaches infinity, the terms with 5/x and 2/x become very close to zero, and we can neglect them:

lim(x->inf) ((1 - 0) / (1 + 0))^(2x+10)

Now, we have:

lim(x->inf) (1)^^(2x+10)

Step 3: Simplify the exponent As x approaches infinity, the exponent 2x+10 also goes to infinity. So we have:

lim(x->inf) (1)^^(inf)

Any nonzero number raised to the power of infinity is still infinity, and 1 is not an exception. Therefore, the final answer is:

lim(x->inf) (1) = 1