Misconception about evaluating limits

hint

$ 1^\infty $ is an inderminate form.

It is equivalent to compute

$$\lim_{x\to 0^+}(\frac 13(3^x+8^x+9^x))^{\frac 1x}$$

or, after taking logarithm,

$$\lim_{x\to 0^+}\frac 1x\ln(1+\frac 13(3^x-1+8^x-1+9^x-1))$$

$$=\lim_{x\to 0}\frac{1}{3x}(3^x-1+8^x-1+9^x-1)$$

Now, use the fact that

$$\lim_{x\to 0^+}\frac{a^x-1}{x}=\ln(a)$$

The result you will find is $e^{\ln(2.3)}=6.$

Tags:

Limits

Calculus

Real Analysis

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