Solution: L'Hopital's Rule #3
Solution: L'Hopital's Rule #3
Evaluate the following limit:
\Large{\lim_{x\to 0}\frac{e^{2x}-\cos(x)}{x}}
Step 1:
Trying to solve this limit by direct substitution will given an indeterminate form. We can use L’hopital’s rule to help us find the limits of indeterminate forms.
For indeterminate forms, L’hopital’s rule tells us:
\large{\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x \to a}\frac{\frac{d}{dx}f(x)}{\frac{d}{dx}g(x)}}
Step 2:
Applying L’hopital’s rule, we have:
\large{\lim_{x \to 0}\frac{e^{2x}-\cos(x)}{x} = \lim_{x \to 0}\frac{\frac{d}{dx}(e^{2x}-\cos(x))}{\frac{d}{dx}(x)}}
Step 3:
Evaluating each of the derivatives individually, we have:
\large{\frac{d}{dx}(e^{2x}-\cos(x)) =2e^{2x}+\sin(x)}
\frac{d}{dx}(x) = 1
Step 4:
Finally, we can solve the limit as follows:
\large{\lim_{x \to 0}\frac{2e^{2x}+\sin(x)}{1} = \frac{2e^{2(0)}+\sin(0)}{1}}
\large{ = \frac{2}{1} = 2}
Thus:
\large{\lim_{x\to 0}\frac{e^{2x}-\cos(x)}{x} = 2}
Solution Complete!
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Solution: L'Hopital's Rule #3
Evaluate the following limit:
\Large{\lim_{x\to 0}\frac{e^{2x}-\cos(x)}{x}}
Step 1:
Trying to solve this limit by direct substitution will given an indeterminate form. We can use L’hopital’s rule to help us find the limits of indeterminate forms.
For indeterminate forms, L’hopital’s rule tells us:
Step 2:
Applying L’hopital’s rule, we have:
Step 3:
Evaluating each of the derivatives individually, we have:
Step 4:
Finally, we can solve the limit as follows:
Thus:
Solution Complete!
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