+118654 \lim _{x\to 2}\left(\frac{\left(2^x-x^2\right)}{\left(x-2\right)}\right)
\(\displaystyle\lim _{x\to 2}\left(\frac{\left(2^x-x^2\right)}{\left(x-2\right)}\right)\\ \text{ I am going to use L'Hopitals rule }\\ \)
First I need to work out the differential of the numerator....
\(Let\;\;y=2^x\\ ln(y)=ln(2^x)\\ \frac{d}{dx}\;ln(y)=\frac{d}{dx}\;ln(2^x)\\ \frac{dy}{dx}\frac{1}{y}=\frac{d}{dx}\: xln(2)\\ \frac{dy}{dx}\frac{1}{y}=ln(2)\\ \frac{dy}{dx}=y*ln(2)\\ \frac{dy}{dx}=2^x\;ln(2)\\\)
so
\(\displaystyle\lim _{x\to 2}\left(\frac{\left(2^x-x^2\right)}{\left(x-2\right)}\right)\\ \displaystyle\lim _{x\to 2}\left(\frac{2^x\;ln(2)-2x}{1}\right)\\ \displaystyle\lim _{x\to 2}\left(2^x\;ln(2)-2x\right)\\ =2^2ln(2)-4\\ =4ln(2)-4\\ \)
Here is the graph:
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Find the following limit:
lim_(x->2) (2^x - x^2)/(x - 2)
Applying l'Hôpital's rule, we get that
lim_(x->2) (2^x - x^2)/(x - 2) | = | lim_(x->2) ( d/( dx)(2^x - x^2))/( d/( dx)(x - 2))
| = | lim_(x->2) (2^x log(2) - 2 x)/1
| = | lim_(x->2) (2^x log(2) - 2 x)
lim_(x->2) (log(2) 2^x - 2 x)
lim_(x->2) (2^x log(2) - 2 x) = 2^2 log(2) - 2 2 = log(16) - 4:
Answer: |log(16) - 4
