Limit calculation
- Thread starter krys235
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- Nov 24, 2012
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Hello, and welcome to FMH! 
I am assuming you are applying L'Hôpital's Rule, and so I would begin by considering the function:
[MATH]y=2^{f(x)}[/MATH]
Taking the natural log of both sides, we obtain:
[MATH]\log(y)=\log(2)f(x)[/MATH]
Implicitly differentiating with respect to \(x\) we obtain:
[MATH]\frac{1}{y}\cdot\d{y}{x}=\log(2)f'(x)[/MATH]
[MATH]\d{y}{x}=y\log(2)f'(x)=\log(2)2^{f(x)}f'(x)[/MATH]
Now, let's consider the given limit:
[MATH]L=\lim_{x\to2}\left(\frac{2^{2^x}-2^{x^2}}{2^x-x^2}\right)[/MATH]
Applying L'Hôpital's Rule, we may state:
[MATH]L=\lim_{x\to2}\left(\frac{\log(2)2^{2^x}\log(2)2^x-\log(2)2^{x^2}(2x)}{\log(2)2^x-2x}\right)[/MATH]
Can you finish?
I am assuming you are applying L'Hôpital's Rule, and so I would begin by considering the function:
[MATH]y=2^{f(x)}[/MATH]
Taking the natural log of both sides, we obtain:
[MATH]\log(y)=\log(2)f(x)[/MATH]
Implicitly differentiating with respect to \(x\) we obtain:
[MATH]\frac{1}{y}\cdot\d{y}{x}=\log(2)f'(x)[/MATH]
[MATH]\d{y}{x}=y\log(2)f'(x)=\log(2)2^{f(x)}f'(x)[/MATH]
Now, let's consider the given limit:
[MATH]L=\lim_{x\to2}\left(\frac{2^{2^x}-2^{x^2}}{2^x-x^2}\right)[/MATH]
Applying L'Hôpital's Rule, we may state:
[MATH]L=\lim_{x\to2}\left(\frac{\log(2)2^{2^x}\log(2)2^x-\log(2)2^{x^2}(2x)}{\log(2)2^x-2x}\right)[/MATH]
Can you finish?