0
$\begingroup$

$$\lim_{x\to 14} \left(\left(x^2-30\cdot x+225\right)^\frac{1}{x^2-29\cdot x +210}\right)$$

I've tried to simply this equation:

$$\lim_{x\to 14} \left(\left(\left(x-15\right)^2\right)^\frac{1}{\left(x-15\right)\left(x-14\right)}\right)$$

$$\lim_{x\to 14} \left(\exp\left(\frac{\ln\left(\left(x-15\right)^2\right)}{\left(x-15\right)\left(x-14\right)}\right)\right)$$

If I substitute $x$ for $14$, then I still end up with division by $0$, which would imply that the limit is undefined, but I think that I miss some further algebraic manipulations here to reduce the expression even further. What can I do here?

$\endgroup$
  • $\begingroup$ Usually what you do is check to see what happens when you approach the limit point from the left and right. It could be the case that both the left and right limits inside of $\exp$ go to $+\infty$ or both go to $-\infty$ (which then would tell you if the limit exists). $\endgroup$ – Cameron Williams Jan 14 '16 at 2:35
2
$\begingroup$

Assume the limit exists and is equal to $L$. Consider

$$\log{L} = \lim_{x \to 14} \frac1{(x-15)(x-14)} \log{(x-15)^2} = \lim_{x \to 0} \frac{2}{x (x-1)} \log{(1-x)} $$

$\endgroup$
  • $\begingroup$ clear and simple +1. $\endgroup$ – Paramanand Singh Jan 14 '16 at 4:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.