Find Limit of $\displaystyle\frac{\sec x +3}{7x-\tan y}$ at (0,$\dfrac{\pi}{4}$)

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Discussion Overview

The discussion revolves around finding the limit of the expression $\displaystyle\frac{\sec x +3}{7x-\tan y}$ as the point $(x,y)$ approaches $(0,\frac{\pi}{4})$. The focus includes the interpretation of limits in multivariable calculus and the conditions under which direct substitution is valid.

Discussion Character

  • Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests that the limit can be evaluated by directly substituting the values of $x$ and $y$ into the expression.
  • Another participant counters that direct substitution may lead to indeterminate forms such as 0 in the denominator or infinity over infinity, indicating potential difficulties in evaluating the limit.
  • A third participant references the definition of continuity, stating that a limit can be evaluated by substitution if the function is continuous at the point in question.

Areas of Agreement / Disagreement

Participants express differing views on the validity of direct substitution for evaluating the limit, indicating that the discussion remains unresolved regarding the best approach to take.

Contextual Notes

Participants highlight potential issues with indeterminate forms when substituting values directly, but do not resolve the implications of continuity on the limit evaluation.

karush
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Find the limit
$\displaystyle\lim_{(x,y) \to \left[0,\dfrac{\pi}{4}\right]}
\dfrac{\sec x +3}{7x-\tan y}= $

I haven't seen limit displayed like this so assume the (x,y) values are just pluged in as first step
 
Last edited:
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No.

$$(x,y) \to \left [0, \dfrac \pi 4\right]$$

is the same as

$$x \to 0,~y \to \dfrac \pi 4$$

i.e the limit at the point $$\left(0,~ \dfrac \pi 4\right)$$

If you can plug them in and get a value, great, that's your limit.
But generally plugging the values in will result in 0 in the denominator, or infinity divided by infinity, or
any of the usual difficulties one encounters doing limit problems.

In this particular problem you can just plug the values in and obtain the limit value directly.
 
$\lim{x\to a} f(x)$ is the same as f(a) if and only if f is continuous at x= a. Indeed that is the definition of "continuous".
 

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