diff --git a/README.md b/README.md index 78a0fc1..00773d9 100644 --- a/README.md +++ b/README.md @@ -90,8 +90,46 @@ This result is obtained because after canceling the common factor, we are left w $$lim_{{\Large \to 1}} \frac{{x^2 - 4x + 3}}{{x - 1}}$$n -
+```latex +\[ +\lim_{{x \to 1}} \frac{{x^2 - 4x + 3}}{{x - 1}} +\] +``` +This will render as: + +$$ +\lim_{{x \to 1}} \frac{{x^2 - 4x + 3}}{{x - 1}} +$$ + +[Solution](0 + +To solve the limit, we can factor the numerator: +$$ +x^2 - 4x + 3 = (x - 1)(x - 3) +$$ + +So the limit becomes: +$$ +\lim_{{x \to 1}} \frac{{(x - 1)(x - 3)}}{{x - 1}} +$$ + +We can cancel out the \((x - 1)\) terms: +$$ +\lim_{{x \to 1}} (x - 3) +$$ + +Now, we can directly substitute \( x = 1 \): +$$ +1 - 3 = -2 +$$ + +Therefore, the limit is: +$$ +\lim_{{x \to 1}} \frac{{x^2 - 4x + 3}}{{x - 1}} = -2 +$$ + +# ### 1d) **Limit Expression:**