{"id":40245,"date":"2025-06-27T15:01:43","date_gmt":"2025-06-27T15:01:43","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=40245"},"modified":"2025-06-27T15:01:44","modified_gmt":"2025-06-27T15:01:44","slug":"the-limit-below-represents-a-derivative","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-limit-below-represents-a-derivative\/","title":{"rendered":"The limit below represents a derivative\u00a0"},"content":{"rendered":"\n
\"\"<\/figure>\n\n\n\n

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

The correct answers are:<\/p>\n\n\n\n

f(x)=x3f(x) = x^3 a=6a = 6<\/p>\n\n\n\n

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

The limit given in the problem is<\/p>\n\n\n\n

lim\u2061h\u2192063+h\u2212216h\\lim_{h \\to 0} \\frac{6^3 + h – 216}{h}<\/p>\n\n\n\n

This expression resembles the definition of the derivative, which is<\/p>\n\n\n\n

f\u2032(a)=lim\u2061h\u21920f(a+h)\u2212f(a)hf'(a) = \\lim_{h \\to 0} \\frac{f(a + h) – f(a)}{h}<\/p>\n\n\n\n

To compare the two expressions, consider the numerator in the problem: 63+h\u22122166^3 + h – 216. This can be rewritten as (216+h\u2212216)=h(216 + h – 216) = h, which simplifies the entire limit to<\/p>\n\n\n\n

lim\u2061h\u21920hh=1\\lim_{h \\to 0} \\frac{h}{h} = 1<\/p>\n\n\n\n

That tells us the slope of the tangent line at a specific point is 1. But the real purpose of this question is to identify which function and which point aa generated this derivative limit.<\/p>\n\n\n\n

If we suppose that f(x)=x3f(x) = x^3, then the derivative is<\/p>\n\n\n\n

f\u2032(x)=3x2f'(x) = 3x^2<\/p>\n\n\n\n

Choosing a=6a = 6, we get<\/p>\n\n\n\n

f\u2032(6)=3\u22c562=108f'(6) = 3 \\cdot 6^2 = 108<\/p>\n\n\n\n

However, this is a mismatch since our limit evaluates to 1, not 108. That discrepancy means we have misread the structure.<\/p>\n\n\n\n

Now focus instead on the functional expression given directly:<\/p>\n\n\n\n

63+h\u2212216h\\frac{6^3 + h – 216}{h}<\/p>\n\n\n\n

Since 63=2166^3 = 216, we see that this is simply<\/p>\n\n\n\n

216+h\u2212216h=hh=1\\frac{216 + h – 216}{h} = \\frac{h}{h} = 1<\/p>\n\n\n\n

Thus, the given expression is<\/p>\n\n\n\n

lim\u2061h\u21920f(a+h)\u2212f(a)h\\lim_{h \\to 0} \\frac{f(a+h) – f(a)}{h}<\/p>\n\n\n\n

with<\/p>\n\n\n\n

f(x)=xf(x) = x<\/p>\n\n\n\n

and<\/p>\n\n\n\n

a=216a = 216<\/p>\n\n\n\n

But this contradicts the original problem’s format, which likely contains a typographical error.<\/p>\n\n\n\n

Looking again, it seems the intended expression may have been<\/p>\n\n\n\n

lim\u2061h\u21920(6+h)3\u2212216h\\lim_{h \\to 0} \\frac{(6 + h)^3 – 216}{h}<\/p>\n\n\n\n

In that case, we would expand (6+h)3=216+108h+18h2+h3(6 + h)^3 = 216 + 108h + 18h^2 + h^3, subtract 216, and the limit becomes<\/p>\n\n\n\n

lim\u2061h\u21920108h+18h2+h3h=lim\u2061h\u21920(108+18h+h2)=108\\lim_{h \\to 0} \\frac{108h + 18h^2 + h^3}{h} = \\lim_{h \\to 0} (108 + 18h + h^2) = 108<\/p>\n\n\n\n

That confirms the function is f(x)=x3f(x) = x^3 and a=6a = 6, because the derivative f\u2032(x)=3x2f'(x) = 3x^2 gives f\u2032(6)=108f'(6) = 108.<\/p>\n\n\n\n

Therefore, the original function must be f(x)=x3f(x) = x^3 and the value of aa is 6<\/strong>.<\/p>\n\n\n\n

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

The Correct Answer and Explanation is: The correct answers are: f(x)=x3f(x) = x^3 a=6a = 6 Explanation: The limit given in the problem is lim\u2061h\u2192063+h\u2212216h\\lim_{h \\to 0} \\frac{6^3 + h – 216}{h} This expression resembles the definition of the derivative, which is f\u2032(a)=lim\u2061h\u21920f(a+h)\u2212f(a)hf'(a) = \\lim_{h \\to 0} \\frac{f(a + h) – f(a)}{h} To compare the […]<\/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-40245","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40245","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=40245"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40245\/revisions"}],"predecessor-version":[{"id":40248,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40245\/revisions\/40248"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=40245"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=40245"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=40245"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}