Limit as $x$ goes to $2$ of $\dfrac{2^x - x^2}{x-2}$

Question

$\displaystyle\lim_{x\to 2}\dfrac{2^x - x^2}{x-2}$

Answer 1

Let $x-2=h$

$$\lim_{h\to0}\dfrac{2^{h+2}-(h+2)^2}h=2^2\lim_{h\to0}\dfrac{2^h-1}h-\lim_{h\to0}(h+2)$$

Now for $a>0, a=e^{\ln a}$

$$\lim_{h\to0}\dfrac{a^h-1}h=\ln a\lim_{h\to0}\dfrac{e^{h\ln a}-1}{h\ln a}=\ln a$$

Answer 2

Herein, we present a way forward that does not use differential calculus, but rather relies only on a standard set of inequalities and the squeeze theorem. To that end, we begin with a primer.

PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}} \tag 1$$

for $x<1$.

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Next, we note that we can write $2^x=4e^{\log(2)(x-2)}$. Thne, using $(2)$ reveals that

$$4(1+\log(2)(x-2)) \le2^x\le \frac{4}{1-\log(2)(x-2)}$$

for $x<2+1/\log(2)$.

Therefore, we can write for $2<x<2+1/\log(2)$

$$4\log(2)-(x+2) \le \frac{2^x-x^2}{x-2}\le \frac{\log(2)x^2 -(x+2)}{1-\log(2)(x-2)} \tag 2$$

while for $x<2$ we can write

$$ \frac{\log(2)x^2 -(x+2)}{1-\log(2)(x-2)} \le \frac{2^x-x^2}{x-2}\le 4\log(2)-(x+2) \tag 3$$

Applying the squeeze theorem to $(2)$ and $(3)$ yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 2}\frac{2^x-x^2}{x-2}=4\log(2)-4}$$