{"id":26940,"date":"2025-06-19T19:51:27","date_gmt":"2025-06-19T19:51:27","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=26940"},"modified":"2025-06-19T19:51:29","modified_gmt":"2025-06-19T19:51:29","slug":"x2-3x-28-lim-x-4-x-4-x2-x-2-lim-x-2","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/x2-3x-28-lim-x-4-x-4-x2-x-2-lim-x-2\/","title":{"rendered":"x2 + 3x = 28 lim X ~4 X – 4 X2 X – 2 lim X -2"},"content":{"rendered":"\n

x2 + 3x = 28 lim X ~4 X – 4 X2 X – 2 lim X -2<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/p>\n\n\n\n

1.<\/strong>lim\u2061x\u21924×2+3x\u221228x\u22124\\lim_{x \\to 4} \\frac{x^2 + 3x – 28}{x – 4}x\u21924lim\u200bx\u22124×2+3x\u221228\u200b<\/p>\n\n\n\n

Step-by-step:<\/strong><\/p>\n\n\n\n

First, factor the numerator:x2+3x\u221228=(x+7)(x\u22124)x^2 + 3x – 28 = (x + 7)(x – 4)x2+3x\u221228=(x+7)(x\u22124)<\/p>\n\n\n\n

So the expression becomes:(x+7)(x\u22124)x\u22124\\frac{(x + 7)(x – 4)}{x – 4}x\u22124(x+7)(x\u22124)\u200b<\/p>\n\n\n\n

Now cancel out the common factor (x\u22124)(x – 4)(x\u22124):(x+7)(x\u22124)(x\u22124)=x+7\\frac{(x + 7)\\cancel{(x – 4)}}{\\cancel{(x – 4)}} = x + 7(x\u22124)\u200b(x+7)(x\u22124)\u200b\u200b=x+7<\/p>\n\n\n\n

Now take the limit as x\u21924x \\to 4x\u21924:lim\u2061x\u21924(x+7)=4+7=11\\lim_{x \\to 4} (x + 7) = 4 + 7 = 11x\u21924lim\u200b(x+7)=4+7=11<\/p>\n\n\n\n

Answer:<\/strong> 11\\boxed{11}11\u200b<\/p>\n\n\n\n

\n\n\n\n

2.<\/strong>lim\u2061x\u2192214\u22121x2x\u22122\\lim_{x \\to 2} \\frac{\\frac{1}{4} – \\frac{1}{x^2}}{x – 2}x\u21922lim\u200bx\u2212241\u200b\u2212x21\u200b\u200b<\/p>\n\n\n\n

Step-by-step:<\/strong><\/p>\n\n\n\n

Combine the numerator into one fraction:14\u22121×2=x2\u221244×2\\frac{1}{4} – \\frac{1}{x^2} = \\frac{x^2 – 4}{4x^2}41\u200b\u2212x21\u200b=4x2x2\u22124\u200b<\/p>\n\n\n\n

Now substitute that back into the main expression:x2\u221244x2x\u22122\\frac{\\frac{x^2 – 4}{4x^2}}{x – 2}x\u221224x2x2\u22124\u200b\u200b<\/p>\n\n\n\n

Recall that x2\u22124=(x\u22122)(x+2)x^2 – 4 = (x – 2)(x + 2)x2\u22124=(x\u22122)(x+2), so we have:(x\u22122)(x+2)4×2(x\u22122)=(x\u22122)(x+2)4×2(x\u22122)=x+24×2\\frac{(x – 2)(x + 2)}{4x^2(x – 2)} = \\frac{\\cancel{(x – 2)}(x + 2)}{4x^2\\cancel{(x – 2)}} = \\frac{x + 2}{4x^2}4×2(x\u22122)(x\u22122)(x+2)\u200b=4×2(x\u22122)\u200b(x\u22122)\u200b(x+2)\u200b=4x2x+2\u200b<\/p>\n\n\n\n

Now take the limit as x\u21922x \\to 2x\u21922:2+24(2)2=416=14\\frac{2 + 2}{4(2)^2} = \\frac{4}{16} = \\frac{1}{4}4(2)22+2\u200b=164\u200b=41\u200b<\/p>\n\n\n\n

Answer:<\/strong> 14\\boxed{\\frac{1}{4}}41\u200b\u200b<\/p>\n\n\n\n

\n\n\n\n

Explanation<\/strong><\/p>\n\n\n\n

These problems test our understanding of evaluating limits, especially when direct substitution gives indeterminate forms like 00\\frac{0}{0}00\u200b. When that happens, we must simplify the expression.<\/p>\n\n\n\n

In the first problem, the limit is:lim\u2061x\u21924×2+3x\u221228x\u22124\\lim_{x \\to 4} \\frac{x^2 + 3x – 28}{x – 4}x\u21924lim\u200bx\u22124×2+3x\u221228\u200b<\/p>\n\n\n\n

Substituting x=4x = 4x=4 into the numerator gives 16+12\u221228=016 + 12 – 28 = 016+12\u221228=0, and the denominator is also 0, giving an indeterminate form. To resolve this, we factor the numerator. Factoring allows cancellation of the troublesome (x\u22124)(x – 4)(x\u22124) term, simplifying the expression to x+7x + 7x+7, which is continuous at x=4x = 4x=4. Evaluating the limit becomes a matter of simple substitution, resulting in 11.<\/p>\n\n\n\n

In the second problem, the limit is:lim\u2061x\u2192214\u22121x2x\u22122\\lim_{x \\to 2} \\frac{\\frac{1}{4} – \\frac{1}{x^2}}{x – 2}x\u21922lim\u200bx\u2212241\u200b\u2212x21\u200b\u200b<\/p>\n\n\n\n

Again, direct substitution gives 14\u221214=0\\frac{1}{4} – \\frac{1}{4} = 041\u200b\u221241\u200b=0 in the numerator and x\u22122=0x – 2 = 0x\u22122=0 in the denominator. This 00\\frac{0}{0}00\u200b form signals the need for algebraic simplification. By combining the fractions in the numerator and factoring the resulting expression, we cancel the (x\u22122)(x – 2)(x\u22122) term. The simplified form is continuous at x=2x = 2x=2, allowing us to directly evaluate the limit and find it equals 14\\frac{1}{4}41\u200b.<\/p>\n\n\n\n

In both cases, factoring and simplification help eliminate indeterminate forms, making it possible to evaluate the limit accurately.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

x2 + 3x = 28 lim X ~4 X – 4 X2 X – 2 lim X -2 The Correct Answer and Explanation is: 1.lim\u2061x\u21924×2+3x\u221228x\u22124\\lim_{x \\to 4} \\frac{x^2 + 3x – 28}{x – 4}x\u21924lim\u200bx\u22124×2+3x\u221228\u200b Step-by-step: First, factor the numerator:x2+3x\u221228=(x+7)(x\u22124)x^2 + 3x – 28 = (x + 7)(x – 4)x2+3x\u221228=(x+7)(x\u22124) So the expression becomes:(x+7)(x\u22124)x\u22124\\frac{(x + 7)(x […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-26940","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26940","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=26940"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26940\/revisions"}],"predecessor-version":[{"id":26946,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/26940\/revisions\/26946"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=26940"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=26940"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=26940"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}