Final answer:
To find the limit of lim x→0 (e⁷x - 1 - 7x) / x², we can use L'Hospital's rule. Applying L'Hospital's rule twice, we find that the limit is 49/2.
Step-by-step explanation:
To find the limit of lim x→0 (e⁷x - 1 - 7x) / x², we can use L'Hospital's rule since it is an indeterminate form of the type 0/0. Applying L'Hospital's rule, we differentiate the numerator and denominator with respect to x. The derivative of e⁷x is 7e⁷x, and the derivative of -7x is -7.
Now, we have the limit of lim x→0 (7e⁷x - 7) / 2x. Evaluating this limit as x approaches 0 gives us (7(1) - 7) / 0, which simplifies to 0 / 0.
Since this new expression is still an indeterminate form, we can apply L'Hospital's rule again. Differentiating the numerator and denominator once more, we get the limit of lim x→0 (49e⁷x) / 2. Substituting x = 0 into this expression gives us (49(1)) / 2 = 49 / 2.